Not in G2 · G3 Additional Maths only

Chapter 17 · Additional Maths · Proofs in plane geometry

Proofs in plane geometry

A proof is a short chain of statements, and every statement has a reason you can name. This chapter shows you how to set one out, how to choose between congruent and similar triangles, and how to use two theorems that are new in Additional Maths: the midpoint theorem and the tangent-chord (alternate segment) theorem. You will also use the circle properties you already know, but only after you have checked the condition that goes with each one.

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Figure 1 · the tangent-chord theorem: the angle between the tangent and the chord equals the angle in the opposite segment

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Reasons win marks

Every line of a proof ends with its reason. Name the angles with three letters and say which theorem: alternate angles, same segment, angle in a semicircle.

Check the condition

A circle theorem needs its condition: the centre for the double angle, a diameter for the right angle, four points on a circle for the cyclic quadrilateral. No condition, no theorem.

Match the vertices

Write congruent and similar triangles with the vertices in matching order, so $\triangle A B C \sim \triangle A {D} E$ tells you which side matches which.

By the end of this chapter

Set out a proof as Given, Reasons and Conclusion, naming the property that justifies each step, including angle facts about parallel lines, triangles and isosceles triangles.
Prove triangles congruent (SSS, SAS, ASA or AAS, RHS) or similar (AA, sides in proportion, or two sides in proportion with the included angle), writing the vertices in matching order, and use the result to prove equal lengths, parallel lines, and products of lengths.
State and use the midpoint theorem and its converse, including ratios of lengths along a chord.
Use the circle angle properties and the tangent-chord theorem, checking the condition of each (centre, same segment, diameter, cyclic) before using it.
Prove that a quadrilateral is a parallelogram or rhombus, that four points lie on a circle, that a line bisects an angle or that two lines are parallel, and combine a geometry proof with a little trigonometry or coordinate geometry of a circle.

✶

Start here

Maths In Real Life

Two real places where a proof pays off: a triangular park whose paths join the midpoints of the sides, and a roundabout with two straight roads touching it. The pictures come first; every angle in each solution is checked on the same figure.

$A plan of a triangular park ABC. D is the midpoint of the side AB and E is the midpoint of the side AC, and a straight path joins D to E.$

A

B

C

D

E

DE

Maths In Real Life 1

Paths across a triangular park

In a triangle, the line joining the midpoints of two sides is parallel to the third side and half as long. A straight path that joins the middle of two park edges is therefore parallel to the third edge.

Not in G2 (G3 Additional Maths only): A triangular park $A B C$ has a path $DE$ joining ${D}$ , the midpoint of the edge $AB$ , to $E$ , the midpoint of the edge $AC$ . The corner $\angle A B C = 68^\circ$ .

(a)Prove that the path $DE$ is parallel to the edge $BC$ .
(b)Find $\angle A {D} E$ .
(c)Write down the ratio of the length of the path $DE$ to the length of the edge $BC$ .

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(a)

Given:

\text{}{D}\text{ is the midpoint of }AB

Given:

\text{E is the midpoint of }AC

DE \parallel BC

(midpoint theorem)

Midpoint theorem in triangle

A B C

(b)

Given:

\text{}{D}\text{ is the midpoint of }AB

Given:

\text{E is the midpoint of }AC

DE \parallel BC

from part (a).

Given:

\angle A B C = 68^\circ

\angle A {D} E = \angle A B C

(corresponding angles, parallel lines), so

\angle A {D} E = 68^\circ

DE \parallel BC

: corresponding angles.

(c)

Given:

\text{D is the midpoint of }AB

Given:

\text{E is the midpoint of }AC

DE = \dfrac{1}{2} BC

(midpoint theorem)

The midpoint theorem gives half the length.

DE : BC = 1 : 2

Half the length means a ratio of

1 : 2

Answer

(a)

DE \parallel BC\ \text{(shown)}

(b)

\angle A {D} E = 68^\circ

(c)

1 : 2

$A plan of a round roundabout. Two straight roads leave the junction P and each touches the roundabout at one point, A on the upper side and B on the lower side. The centre O of the roundabout is marked, with dashed radii to A and to B.$

O

A

B

P

OA

OB

PA

PB

Maths In Real Life 2

Two roads that touch a roundabout

A road that just touches a circular roundabout, without entering it, is a tangent to the circle. The two roads that leave a junction and touch the roundabout have the same length from the junction to the point of contact.

Not in G2 (G3 Additional Maths only): A circular roundabout has centre $O$ , and $A$ and $B$ are on its edge. Two straight roads meet at the junction $P$ , and each touches the roundabout at one point: the road $PA$ touches it at $A$ and the road $PB$ touches it at $B$ . A road is perpendicular to the radius at its point of contact: $\angle O A P = 90^\circ$ and $\angle O B P = 90^\circ$ . The angle at the centre is $\angle A O B = 80^\circ$ .

(a)Prove that the two roads $PA$ and $PB$ have the same length.
(b)Find the angle $\angle A P B$ between the two roads.

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Colour keythe angle or statement you were asked for

(a)

The road

PA

touches the roundabout at

A

, so it is perpendicular to the radius

OA

\angle O A P = 90^\circ

A road that touches the circle is a tangent: it meets the radius at a right angle.

The road

PB

touches the roundabout at

B

, so

\angle O B P = 90^\circ

The same fact for the second road.

OA = OB

(radii of the roundabout).

A

and

B

are on the edge of the roundabout.

OP = OP

(common side).

OP

belongs to both triangles.

Each triangle has a right angle, the common hypotenuse

OP

and

OA = OB

, so

\triangle O A P \equiv \triangle O B P

(RHS).

RHS: a right angle, the hypotenuse and one other side.

The matching sides are equal, so

PA = PB

PA

matches

PB

in the congruent triangles.

(b)

Given:

\angle O A P = 90^\circ

Given:

\angle O B P = 90^\circ

\angle O A P + \angle O B P = 180^\circ

, and these are opposite angles of

O A P B

, so

O

A

P

B

lie on a circle.

\angle A O B + \angle A P B = 180^\circ

(opposite angles of a cyclic quadrilateral). So

\angle A P B

is:

\displaystyle 180^\circ - 80^\circ = \textcolor{#010106}{100^\circ}

Opposite angles of the cyclic quadrilateral

O A P B

Answer

(a)

PA = PB\ \text{(shown)}

(b)

\angle A P B = 100^\circ

Supports K341-G3-01

Writing a proof: lines, triangles and reasons

Not in G2 (G3 Additional Maths only): writing proofs in plane geometry is G3 Additional Maths only. Everything in this section is a fact from earlier years; the new skill is writing it as a proof with a reason for every step.

A proof is a short chain of statements. Each statement is either given in the question or follows from earlier ones, and beside each you write its reason: the name of the property that makes it true. A reader should be able to check every line without looking at the figure.

Revision. Parallel lines: corresponding angles are equal, alternate angles are equal and co-interior angles add up to

180^\circ

. Triangles: the angles add up to

180^\circ

, an exterior angle equals the sum of the two interior opposite angles, and the base angles of an isosceles triangle are equal. Straight lines: angles on a line add up to

180^\circ

, angles at a point add up to

360^\circ

and vertically opposite angles are equal. Parallelogram: both pairs of opposite sides are parallel, and its diagonals bisect each other.

The Core Idea

Write a proof as Given, Reasons, Conclusion.

State what you are given. Then write each step as a claim with its reason in brackets, and finish by stating the conclusion the question asked for. A step with no reason earns no mark, and a reason must name the property, not the picture: say alternate angles, not “they look equal”.

SGSchoolKaki Tip

Name every angle with three letters. Write $\angle P R U$ , not “angle $R$ ”, because several angles meet at $R$ . The middle letter is the vertex. A reason such as “alternate angles” then points at two named angles, and the reader can check it.

In the first Example every line of the solution has the same layout: a claim, the reason in brackets, and what it tells you. The first answer uses alternate angles; the second uses corresponding angles and then a straight line.

Example 1.1

Parallel lines and a transversal

$PQ$ and $US$ are parallel lines. A straight line $VX$ cuts them at $P$ and $R$ , with $V$ , $P$ , $R$ , $X$ in that order along it. $Q$ and $S$ are on the same side of $VX$ , and $U$ is on the other side. $\angle Q P R = 52^\circ$ . Find $x$ , the size of $\angle P R U$ , and $y$ , the size of $\angle P R S$ .

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Colour keythe angle or statement you were asked for

Given:

PQ \parallel US

The stem says the lines are parallel, so the angle rules for parallel lines may be used.

Given:

\angle Q P R = 52^\circ

\angle P R U = \angle Q P R

(alternate angles, parallel lines), so

\angle P R U = 52^\circ

\angle P R U

and

\angle Q P R

are on opposite sides of the transversal, between the parallel lines.

\angle S R X = \angle Q P R

(corresponding angles, parallel lines), so

\angle S R X = 52^\circ

\angle S R X

sits at

R

exactly where

\angle Q P R

sits at

P

\angle S R X + \angle P R S = 180^\circ

(angles on a straight line add up to

180^\circ

). So

\angle P R S

is:

\displaystyle 180^\circ - 52^\circ = \textcolor{#010106}{128^\circ}

P

R

and

X

are on one straight line, so the two angles fill a half-turn.

Answer

\angle P R U = 52^\circ

;

\angle P R S = 128^\circ

Your Turn 1.1

$PQ$ and $US$ are parallel lines. A straight line $VX$ cuts them at $P$ and $R$ , with $V$ , $P$ , $R$ , $X$ in that order along it. $Q$ and $S$ are on the same side of $VX$ , and $U$ is on the other side. $\angle Q P R = 64^\circ$ . Find $x$ , the size of $\angle P R U$ , and $y$ , the size of $\angle P R S$ .

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Colour keythe angle or statement you were asked for

Given:

PQ \parallel US

The stem says the lines are parallel, so the angle rules for parallel lines may be used.

Given:

\angle Q P R = 64^\circ

\angle P R U = \angle Q P R

(alternate angles, parallel lines), so

\angle P R U = 64^\circ

\angle P R U

and

\angle Q P R

are on opposite sides of the transversal, between the parallel lines.

\angle S R X = \angle Q P R

(corresponding angles, parallel lines), so

\angle S R X = 64^\circ

\angle S R X

sits at

R

exactly where

\angle Q P R

sits at

P

\angle S R X + \angle P R S = 180^\circ

(angles on a straight line add up to

180^\circ

). So

\angle P R S

is:

\displaystyle 180^\circ - 64^\circ = \textcolor{#010106}{116^\circ}

P

R

and

X

are on one straight line, so the two angles fill a half-turn.

Answer

\angle P R U = 64^\circ

;

\angle P R S = 116^\circ

Which pair of angles you name decides the reason. The table shows the three pairs that parallel lines make with a transversal; the marked angles are the pair in each row.

The three angle pairs made by parallel lines
Pair of angles	What the pair satisfies	Picture
Corresponding angles: same position at the two crossings	equal	corresponding pair
Alternate angles: opposite sides of the transversal, between the lines	equal	alternate pair
Co-interior angles: same side of the transversal, between the lines	add up to $180^\circ$	co-interior pair

Think About It

A line cuts two lines at

P

and

R

. The alternate angles it makes measure

52^\circ

and

61^\circ

. A student writes “the two lines are parallel, because the angles look about the same”. Is the conclusion proved?

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Parallel lines make equal alternate angles, so look at whether the two angles are equal.

\displaystyle 61^\circ - 52^\circ = 9^\circ

The difference of the two alternate angles: it is not zero, so they are not equal.

Answer

No. The alternate angles differ by

9^\circ

, so they are not equal and the lines are not parallel. “Looks about the same” is never a reason: a proof needs a property that is true, and each step must name it.

The exterior angle theorem is not a new fact: it is proved from two facts you already know, the angle sum of a triangle and angles on a straight line. The next Example uses it and then proves it.

Example 1.2

Angle sum and the exterior angle

In triangle $A B C$ , $\angle C A B = 48^\circ$ and $\angle A B C = 64^\circ$ . The side $AB$ is extended to ${D}$ . (a) Find $x$ , the size of $\angle A C B$ . (b) Prove that the exterior angle $\angle C B {D}$ equals $\angle C A B + \angle A C B$ .

Show solutionHide solution

Colour keythe angle or statement you were asked for

Given:

\angle C A B = 48^\circ

Given:

\angle A B C = 64^\circ

\angle C A B + \angle A B C + \angle A C B = 180^\circ

(angle sum of a triangle). So

\angle A C B

is:

\displaystyle 180^\circ - 48^\circ - 64^\circ = \textcolor{#010106}{68^\circ}

The three angles of triangle

A B C

add up to a half-turn.

\angle A B C + \angle C B {D} = 180^\circ

(angles on a straight line add up to

180^\circ

). So

\angle C B {D}

is:

\displaystyle 180^\circ - 64^\circ = \textcolor{#010106}{116^\circ}

A

B

and

{D}

are on one straight line, so the angles at

B

on that line add up to

180^\circ

\angle C A B + \angle A B C + \angle A C B = 180^\circ

(angle sum of a triangle)

The same angle sum as in (a), now used as a statement, not as a calculation.

Both

\angle C B {D}

and

\angle C A B + \angle A C B

equal

180^\circ - \angle A B C

, so

\angle C B {D} = \angle C A B + \angle A C B

Subtracting

\angle A B C

from

180^\circ

on both lines gives the same angle: that is the exterior angle theorem.

Answer

\angle A C B = 68^\circ

;

\angle C B {D} = \angle C A B + \angle A C B\ \text{(shown)}

Your Turn 1.2

In triangle $A B C$ , $\angle C A B = 58^\circ$ and $\angle A B C = 36^\circ$ . The side $AB$ is extended to ${D}$ . (a) Find $x$ , the size of $\angle A C B$ . (b) Prove that the exterior angle $\angle C B {D}$ equals $\angle C A B + \angle A C B$ .

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Colour keythe angle or statement you were asked for

Given:

\angle C A B = 58^\circ

Given:

\angle A B C = 36^\circ

\angle C A B + \angle A B C + \angle A C B = 180^\circ

(angle sum of a triangle). So

\angle A C B

is:

\displaystyle 180^\circ - 58^\circ - 36^\circ = \textcolor{#010106}{86^\circ}

The three angles of triangle

A B C

add up to a half-turn.

\angle A B C + \angle C B {D} = 180^\circ

(angles on a straight line add up to

180^\circ

). So

\angle C B {D}

is:

\displaystyle 180^\circ - 36^\circ = \textcolor{#010106}{144^\circ}

A

B

and

{D}

are on one straight line, so the angles at

B

on that line add up to

180^\circ

\angle C A B + \angle A B C + \angle A C B = 180^\circ

(angle sum of a triangle)

The same angle sum as in (a), now used as a statement, not as a calculation.

Both

\angle C B {D}

and

\angle C A B + \angle A C B

equal

180^\circ - \angle A B C

, so

\angle C B {D} = \angle C A B + \angle A C B

Subtracting

\angle A B C

from

180^\circ

on both lines gives the same angle: that is the exterior angle theorem.

Answer

\angle A C B = 86^\circ

;

\angle C B {D} = \angle C A B + \angle A C B\ \text{(shown)}

Isosceles triangles give a standard deduction: equal sides give equal base angles, and equal base angles can then be carried to another triangle by parallel lines.

Example 1.3

An isosceles triangle cut by a parallel line

In triangle $A B C$ , $CA = CB$ and $\angle C A B = 70^\circ$ . The point ${D}$ lies on $CA$ and the point $E$ lies on $CB$ so that $DE$ is parallel to $AB$ . (a) Prove that $\angle C {D} E = \angle C E {D}$ . (b) Find $x$ , the size of $\angle A {D} E$ .

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Colour keythe angle or statement you were asked for

Given:

CA = CB

Given:

DE \parallel AB

Given:

\angle C A B = 70^\circ

\angle C B A = \angle C A B

(base angles of an isosceles triangle), so

\angle C B A = 70^\circ

CA = CB

, so the base angles at

A

and

B

are equal.

\angle C {D} E = \angle C A B

(corresponding angles, parallel lines), so

\angle C {D} E = 70^\circ

DE \parallel AB

and the line

CA

crosses both:

\angle C {D} E

and

\angle C A B

are in matching positions.

\angle C E {D} = \angle C B A

(corresponding angles, parallel lines), so

\angle C E {D} = 70^\circ

The line

CB

crosses the same parallel lines:

\angle C E {D}

matches

\angle C B A

\angle C {D} E = \angle C A B = \angle C B A = \angle C E {D}

, so

\angle C {D} E = \angle C E {D}

Both angles are equal to the same base angle, so they are equal to each other.

\angle C {D} E + \angle A {D} E = 180^\circ

(angles on a straight line add up to

180^\circ

). So

\angle A {D} E

is:

\displaystyle 180^\circ - 70^\circ = \textcolor{#010106}{110^\circ}

A

{D}

and

C

are on one straight line, so the angles at

{D}

on it add up to

180^\circ

Answer

\angle C {D} E = \angle C E {D}\ \text{(shown)}

;

\angle A {D} E = 110^\circ

Your Turn 1.3

In triangle $A B C$ , $CA = CB$ and $\angle C A B = 50^\circ$ . The point ${D}$ lies on $CA$ and the point $E$ lies on $CB$ so that $DE$ is parallel to $AB$ . (a) Prove that $\angle C {D} E = \angle C E {D}$ . (b) Find $x$ , the size of $\angle A {D} E$ .

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Colour keythe angle or statement you were asked for

Given:

CA = CB

Given:

DE \parallel AB

Given:

\angle C A B = 50^\circ

\angle C B A = \angle C A B

(base angles of an isosceles triangle), so

\angle C B A = 50^\circ

CA = CB

, so the base angles at

A

and

B

are equal.

\angle C {D} E = \angle C A B

(corresponding angles, parallel lines), so

\angle C {D} E = 50^\circ

DE \parallel AB

and the line

CA

crosses both:

\angle C {D} E

and

\angle C A B

are in matching positions.

\angle C E {D} = \angle C B A

(corresponding angles, parallel lines), so

\angle C E {D} = 50^\circ

The line

CB

crosses the same parallel lines:

\angle C E {D}

matches

\angle C B A

\angle C {D} E = \angle C A B = \angle C B A = \angle C E {D}

, so

\angle C {D} E = \angle C E {D}

Both angles are equal to the same base angle, so they are equal to each other.

\angle C {D} E + \angle A {D} E = 180^\circ

(angles on a straight line add up to

180^\circ

). So

\angle A {D} E

is:

\displaystyle 180^\circ - 50^\circ = \textcolor{#010106}{130^\circ}

A

{D}

and

C

are on one straight line, so the angles at

{D}

on it add up to

180^\circ

Answer

\angle C {D} E = \angle C E {D}\ \text{(shown)}

;

\angle A {D} E = 130^\circ

When a line bisects an angle, the two halves are equal. That equality is part of what you are given, so its reason is written as “given” and you may use it at once.

Example 1.4

An angle bisector

In triangle $A B C$ , $AD$ is the bisector of $\angle B A C$ and ${D}$ lies on $BC$ . $\angle B A {D} = 40^\circ$ and $\angle A B {D} = 56^\circ$ . (a) Find $x$ , the size of $\angle A {D} B$ . (b) Show that $\angle A {D} C = \angle {D} A C + \angle A B {D}$ .

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Colour keythe angle or statement you were asked for

Given:

\angle B A {D} = 40^\circ

Given:

\angle A B {D} = 56^\circ

\angle B A {D} + \angle A B {D} + \angle A {D} B = 180^\circ

(angle sum of a triangle). So

\angle A {D} B

is:

\displaystyle 180^\circ - 40^\circ - 56^\circ = \textcolor{#010106}{84^\circ}

The three angles of triangle

A B {D}

add up to a half-turn.

Given:

\angle B A {D} = \angle {D} A C

AD

bisects

\angle B A C

, so it splits the angle into two equal angles: this is the information given.

\angle A {D} B + \angle A {D} C = 180^\circ

(angles on a straight line add up to

180^\circ

). So

\angle A {D} C

is:

\displaystyle 180^\circ - 84^\circ = \textcolor{#010106}{96^\circ}

B

{D}

and

C

are on one straight line, so the angles at

{D}

add up to

180^\circ

\angle A {D} C = \angle B A {D} + \angle A B {D}

(exterior angle of a triangle)

\angle A {D} C

is the exterior angle of triangle

A B {D}

{D}

, so it equals the two interior opposite angles.

Replace

\angle B A {D}

by the equal angle

\angle {D} A C

(given):

\angle A {D} C = \angle {D} A C + \angle A B {D}

. So

\angle {D} A C

is:

\displaystyle 96^\circ - 56^\circ = \textcolor{#010106}{40^\circ}

The bisector lets one half of the angle stand in for the other; the difference below is the angle

{D} A C

, equal to

\angle B A {D}

Answer

\angle A {D} B = 84^\circ

;

\angle A {D} C = \angle {D} A C + \angle A B {D}\ \text{(shown)}

Your Turn 1.4

In triangle $A B C$ , $AD$ is the bisector of $\angle B A C$ and ${D}$ lies on $BC$ . $\angle B A {D} = 30^\circ$ and $\angle A B {D} = 70^\circ$ . (a) Find $x$ , the size of $\angle A {D} B$ . (b) Show that $\angle A {D} C = \angle {D} A C + \angle A B {D}$ .

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Colour keythe angle or statement you were asked for

Given:

\angle B A {D} = 30^\circ

Given:

\angle A B {D} = 70^\circ

\angle B A {D} + \angle A B {D} + \angle A {D} B = 180^\circ

(angle sum of a triangle). So

\angle A {D} B

is:

\displaystyle 180^\circ - 30^\circ - 70^\circ = \textcolor{#010106}{80^\circ}

The three angles of triangle

A B {D}

add up to a half-turn.

Given:

\angle B A {D} = \angle {D} A C

AD

bisects

\angle B A C

, so it splits the angle into two equal angles: this is the information given.

\angle A {D} B + \angle A {D} C = 180^\circ

(angles on a straight line add up to

180^\circ

). So

\angle A {D} C

is:

\displaystyle 180^\circ - 80^\circ = \textcolor{#010106}{100^\circ}

B

{D}

and

C

are on one straight line, so the angles at

{D}

add up to

180^\circ

\angle A {D} C = \angle B A {D} + \angle A B {D}

(exterior angle of a triangle)

\angle A {D} C

is the exterior angle of triangle

A B {D}

{D}

, so it equals the two interior opposite angles.

Replace

\angle B A {D}

by the equal angle

\angle {D} A C

(given):

\angle A {D} C = \angle {D} A C + \angle A B {D}

. So

\angle {D} A C

is:

\displaystyle 100^\circ - 70^\circ = \textcolor{#010106}{30^\circ}

The bisector lets one half of the angle stand in for the other; the difference below is the angle

{D} A C

, equal to

\angle B A {D}

Answer

\angle A {D} B = 80^\circ

;

\angle A {D} C = \angle {D} A C + \angle A B {D}\ \text{(shown)}

When two lines cross, four angles meet at one point. Say which angles each reason links: vertically opposite angles share no arm, and angles on a line share the arm between them.

Example 1.5

Lines that cross

The straight lines $AB$ and $CD$ meet at $X$ , with $X$ between $A$ and $B$ and between $C$ and ${D}$ . $\angle A X C = 70^\circ$ . Find $x$ , the size of $\angle B X {D}$ , $y$ , the size of $\angle C X B$ , and $z$ , the size of $\angle {D} X A$ .

Show solutionHide solution

Colour keythe angle or statement you were asked for

Given:

\angle A X C = 70^\circ

\angle B X {D} = \angle A X C

(vertically opposite angles are equal), so

\angle B X {D} = 70^\circ

XB

is opposite

XA

and

XD

is opposite

XC

, so the angles are vertically opposite.

\angle A X C + \angle C X B = 180^\circ

(angles on a straight line add up to

180^\circ

). So

\angle C X B

is:

\displaystyle 180^\circ - 70^\circ = \textcolor{#010106}{110^\circ}

A

X

and

B

are on one straight line.

\angle A X C + \angle C X B + \angle B X {D} + \angle {D} X A = 360^\circ

(angles at a point add up to

360^\circ

). So

\angle {D} X A

is:

\displaystyle 360^\circ - 70^\circ - 110^\circ - 70^\circ = \textcolor{#010106}{110^\circ}

The four angles fill the whole turn at

X

; the first three are already found.

Answer

\angle B X {D} = 70^\circ

;

\angle C X B = 110^\circ

;

\angle {D} X A = 110^\circ

Your Turn 1.5

The straight lines $AB$ and $CD$ meet at $X$ , with $X$ between $A$ and $B$ and between $C$ and ${D}$ . $\angle A X C = 50^\circ$ . Find $x$ , the size of $\angle B X {D}$ , $y$ , the size of $\angle C X B$ , and $z$ , the size of $\angle {D} X A$ .

Show full solutionHide solution

Colour keythe angle or statement you were asked for

Given:

\angle A X C = 50^\circ

\angle B X {D} = \angle A X C

(vertically opposite angles are equal), so

\angle B X {D} = 50^\circ

XB

is opposite

XA

and

XD

is opposite

XC

, so the angles are vertically opposite.

\angle A X C + \angle C X B = 180^\circ

(angles on a straight line add up to

180^\circ

). So

\angle C X B

is:

\displaystyle 180^\circ - 50^\circ = \textcolor{#010106}{130^\circ}

A

X

and

B

are on one straight line.

\angle A X C + \angle C X B + \angle B X {D} + \angle {D} X A = 360^\circ

(angles at a point add up to

360^\circ

). So

\angle {D} X A

is:

\displaystyle 360^\circ - 50^\circ - 130^\circ - 50^\circ = \textcolor{#010106}{130^\circ}

The four angles fill the whole turn at

X

; the first three are already found.

Answer

\angle B X {D} = 50^\circ

;

\angle C X B = 130^\circ

;

\angle {D} X A = 130^\circ

Supports K341-G3-01

Congruent and similar triangles

Not in G2 (G3 Additional Maths only): proofs in plane geometry are G3 Additional Maths only. You already know congruent and similar triangles from Mathematics, and here you use them as reasons inside a proof.

Revision. Two triangles are congruent when one fits exactly on the other, and similar when they have the same shape but not always the same size. In both, the corresponding angles are equal.

A proof that uses triangles has a fixed pattern. First you state three facts about the two triangles, then you name the test, then you read off the new fact you wanted, such as a pair of equal angles, a pair of equal sides, or a pair of parallel lines. The facts you are given decide which test you use. The table lists every test you may quote.

Which test proves which relation
Test	What you must know	Conclusion
three pairs of equal sides SSS	All three sides of one triangle equal the three sides of the other.	Congruent
two sides and the angle between them SAS	Two sides and the included angle (the angle between them).	Congruent
two angles and a side ASA or AAS	Two angles and a corresponding side: the side between the two angles (ASA) or a side opposite one of them (AAS). The third angle follows from the angle sum.	Congruent
right angle, hypotenuse, one side RHS	A right angle, the hypotenuse and one other side.	Congruent
two pairs of equal angles AA	Two pairs of equal angles (the third pair then follows).	Similar
three pairs of sides in the same ratio Three sides in proportion	All three sides of one triangle are the same multiple of the matching sides of the other: $\dfrac{DE}{AB} = \dfrac{EF}{BC} = \dfrac{FD}{CA}$ .	Similar
two sides in the same ratio and the angle between them Two sides in proportion, included angle equal	Two pairs of sides in the same ratio, $\dfrac{DE}{AB} = \dfrac{DF}{AC}$ , and the included angles equal.	Similar
the same two sides and angle fit two triangles SSA is not a test	Two sides and an angle that is not between them. This does not fix the triangle: the same facts fit two different triangles.	No conclusion

The Core Idea

Choose the test from what you are given, then write the vertices in matching order.

Write $\triangle A B C \equiv \triangle {D} E F$ only when $A$ matches ${D}$ , $B$ matches $E$ and $C$ matches $F$ . The letters tell the reader which sides and angles are equal. After the test, the phrase corresponding parts of congruent triangles (or of similar triangles) gives every other fact.

Example 2.1

Congruent triangles by SSS

The triangle $A B C$ is isosceles with $CA = CB$ , and $M$ is the midpoint of $AB$ , so $AM = BM$ . The angle $\angle C A B = 64^\circ$ . Prove that triangles $A M C$ and $B M C$ are congruent and that $CM$ bisects $\angle A C B$ , then find $\angle A C B$ .

Show solutionHide solution

Colour keythe angle or statement you were asked for

Given:

CA = CB

Given:

AM = BM

CM = CM

(common side)

CM

is a side of both triangles.

\triangle A M C \equiv \triangle B M C

(SSS)

Three pairs of equal sides, matched in the order

A \to B

M \to M

C \to C

\angle A C M = \angle B C M

(corresponding parts of congruent triangles)

Matching angles of congruent triangles are equal, so

CM

splits the angle at

C

into two equal parts.

\angle C B A = \angle C A B

(base angles of an isosceles triangle), so

\angle C B A = 64^\circ

\angle C A B + \angle C B A + \angle A C B = 180^\circ

(angle sum of a triangle). So

\angle A C B

is:

\displaystyle 180^\circ - 64^\circ - 64^\circ = \textcolor{#010106}{52^\circ}

Subtract the known angles from

180^\circ

Answer

\triangle A M C \equiv \triangle B M C\text{ and }\angle A C M = \angle B C M\text{, so } CM \text{ bisects } \angle A C B\ \text{(shown)}

;

\angle A C B = 52^\circ

Your Turn 2.1

The triangle $A B C$ is isosceles with $CA = CB$ , and $M$ is the midpoint of $AB$ , so $AM = BM$ . The angle $\angle C A B = 52^\circ$ . Prove that triangles $A M C$ and $B M C$ are congruent and that $CM$ bisects $\angle A C B$ , then find $\angle A C B$ .

Show full solutionHide solution

Colour keythe angle or statement you were asked for

Given:

CA = CB

Given:

AM = BM

CM = CM

(common side)

CM

is a side of both triangles.

\triangle A M C \equiv \triangle B M C

(SSS)

Three pairs of equal sides, matched in the order

A \to B

M \to M

C \to C

\angle A C M = \angle B C M

(corresponding parts of congruent triangles)

Matching angles of congruent triangles are equal, so

CM

splits the angle at

C

into two equal parts.

\angle C B A = \angle C A B

(base angles of an isosceles triangle), so

\angle C B A = 52^\circ

\angle C A B + \angle C B A + \angle A C B = 180^\circ

(angle sum of a triangle). So

\angle A C B

is:

\displaystyle 180^\circ - 52^\circ - 52^\circ = \textcolor{#010106}{76^\circ}

Subtract the known angles from

180^\circ

Answer

\triangle A M C \equiv \triangle B M C\text{ and }\angle A C M = \angle B C M\text{, so } CM \text{ bisects } \angle A C B\ \text{(shown)}

;

\angle A C B = 76^\circ

Vertex order is part of the statement.

\triangle A M C \equiv \triangle B M C

says

A

matches

B

. Writing

\triangle A M C \equiv \triangle M B C

would claim that

A

matches

M

, which is false even though the same two triangles are meant.

Example 2.2

Congruent triangles by SAS and parallel lines

The straight lines $AB$ and $CD$ cross at $X$ . $X$ is the midpoint of $AB$ and also the midpoint of $CD$ . The angle $\angle A X C = 80^\circ$ . Prove that $AC$ is parallel to $BD$ and find $\angle C X B$ .

Show solutionHide solution

Colour keythe angle or statement you were asked for

Given:

XA = XB

Given:

XC = XD

\angle B X {D} = \angle A X C

(vertically opposite angles are equal), so

\angle B X {D} = 80^\circ

The angle between the equal sides of one triangle matches the angle between the equal sides of the other.

\triangle A X C \equiv \triangle B X {D}

(SAS)

Two sides and the included angle: the equal angle sits between the two equal sides.

\angle C A X = \angle {D} B X

(corresponding parts of congruent triangles)

\angle C A X = \angle {D} B X

, and these are alternate angles on the transversal

AB

, so

AC \parallel BD

(equal alternate angles).

Equal alternate angles are what make two lines parallel.

\angle A X C + \angle C X B = 180^\circ

(angles on a straight line add up to

180^\circ

). So

\angle C X B

is:

\displaystyle 180^\circ - 80^\circ = \textcolor{#010106}{100^\circ}

A

X

B

are on a straight line.

Answer

AC \parallel BD\ \text{(shown)}

;

\angle C X B = 100^\circ

Your Turn 2.2

The straight lines $AB$ and $CD$ cross at $X$ . $X$ is the midpoint of $AB$ and also the midpoint of $CD$ . The angle $\angle A X C = 60^\circ$ . Prove that $AC$ is parallel to $BD$ and find $\angle C X B$ .

Show full solutionHide solution

Colour keythe angle or statement you were asked for

Given:

XA = XB

Given:

XC = XD

\angle B X {D} = \angle A X C

(vertically opposite angles are equal), so

\angle B X {D} = 60^\circ

The angle between the equal sides of one triangle matches the angle between the equal sides of the other.

\triangle A X C \equiv \triangle B X {D}

(SAS)

Two sides and the included angle: the equal angle sits between the two equal sides.

\angle C A X = \angle {D} B X

(corresponding parts of congruent triangles)

\angle C A X = \angle {D} B X

, and these are alternate angles on the transversal

AB

, so

AC \parallel BD

(equal alternate angles).

Equal alternate angles are what make two lines parallel.

\angle A X C + \angle C X B = 180^\circ

(angles on a straight line add up to

180^\circ

). So

\angle C X B

is:

\displaystyle 180^\circ - 60^\circ = \textcolor{#010106}{120^\circ}

A

X

B

are on a straight line.

Answer

AC \parallel BD\ \text{(shown)}

;

\angle C X B = 120^\circ

Think About It

In triangle

A B C

the side

AB = 8

, the side

BC = 5

and the angle

\angle B A C = 30^\circ

. The side

BC

is not between the given angle and the other given side. Is the triangle

A B C

the only triangle with these facts?

Show the full workingHide the working

Put

A

at the origin and draw

AB

along the

x

-axis. The point

C

lies on the ray from

A

that makes

30^\circ

with

AB

, and it is

5

units from

B

Two sides and a non-included angle: the arrangement called SSA, which is not a test.

\displaystyle AC\;= 8\cos 30^\circ + \sqrt{5^2 - (8\sin 30^\circ)^2} = 3 + 4 \sqrt{3}

The circle of radius

5

about

B

cuts the ray at a first point.

\displaystyle AC\;= 8\cos 30^\circ - \sqrt{5^2 - (8\sin 30^\circ)^2} = -3 + 4 \sqrt{3}

...and at a second point, because the square root can take either sign.

Answer

No. The circle cuts the ray twice, so there are two triangles, with

AC = 3 + 4 \sqrt{3}

and

AC = -3 + 4 \sqrt{3}

, and both have

AB = 8

BC = 5

and

\angle B A C = 30^\circ

. Two sides and an angle that is not between them do not make a congruence test, so never write SSA as a reason.

Example 2.3

Similar triangles by AA

In triangle $A B C$ the points ${D}$ and $E$ lie on $AB$ and $AC$ so that $DE \parallel BC$ . The angles are $\angle B A C = 60^\circ$ and $\angle A B C = 70^\circ$ . Prove that triangles $A {D} E$ and $A B C$ are similar and that $AD \times BC = AB \times DE$ , then find $\angle A E {D}$ .

Show solutionHide solution

Colour keythe angle or statement you were asked for

Given:

DE \parallel BC

\angle A {D} E = \angle A B C

(corresponding angles, parallel lines), so

\angle A {D} E = 70^\circ

DE \parallel BC

with transversal

AB

: corresponding angles.

\angle {D} A E = \angle B A C

(common angle), so

\angle {D} A E = 60^\circ

Both angles are at

A

between the same two rays.

\triangle A {D} E \sim \triangle A B C

(AA: two pairs of equal angles)

Two pairs of equal angles. The vertices are in matching order:

A \to A

{D} \to B

E \to C

\dfrac{AD}{AB} = \dfrac{DE}{BC}

(corresponding sides of similar triangles), so

AD \times BC = AB \times DE

Cross-multiply the equal ratios.

\angle A C B + \angle B A C + \angle A B C = 180^\circ

(angle sum of a triangle). So

\angle A C B

is:

\displaystyle 180^\circ - 60^\circ - 70^\circ = \textcolor{#010106}{50^\circ}

Subtract the known angles from

180^\circ

\angle A C B = \angle A E {D}

(corresponding parts of similar triangles), so

\angle A E {D} = 50^\circ

E

matches

C

in the similar triangles.

Answer

\triangle A {D} E \sim \triangle A B C\text{ and }AD \times BC = AB \times DE\ \text{(shown)}

;

\angle A E {D} = 50^\circ

Your Turn 2.3

In triangle $A B C$ the points ${D}$ and $E$ lie on $AB$ and $AC$ so that $DE \parallel BC$ . The angles are $\angle B A C = 40^\circ$ and $\angle A B C = 75^\circ$ . Prove that triangles $A {D} E$ and $A B C$ are similar and that $AD \times BC = AB \times DE$ , then find $\angle A E {D}$ .

Show full solutionHide solution

Colour keythe angle or statement you were asked for

Given:

DE \parallel BC

\angle A {D} E = \angle A B C

(corresponding angles, parallel lines), so

\angle A {D} E = 75^\circ

DE \parallel BC

with transversal

AB

: corresponding angles.

\angle {D} A E = \angle B A C

(common angle), so

\angle {D} A E = 40^\circ

Both angles are at

A

between the same two rays.

\triangle A {D} E \sim \triangle A B C

(AA: two pairs of equal angles)

Two pairs of equal angles. The vertices are in matching order:

A \to A

{D} \to B

E \to C

\dfrac{AD}{AB} = \dfrac{DE}{BC}

(corresponding sides of similar triangles), so

AD \times BC = AB \times DE

Cross-multiply the equal ratios.

\angle A C B + \angle B A C + \angle A B C = 180^\circ

(angle sum of a triangle). So

\angle A C B

is:

\displaystyle 180^\circ - 40^\circ - 75^\circ = \textcolor{#010106}{65^\circ}

Subtract the known angles from

180^\circ

\angle A C B = \angle A E {D}

(corresponding parts of similar triangles), so

\angle A E {D} = 65^\circ

E

matches

C

in the similar triangles.

Answer

\triangle A {D} E \sim \triangle A B C\text{ and }AD \times BC = AB \times DE\ \text{(shown)}

;

\angle A E {D} = 65^\circ

SGSchoolKaki Tip

To prove a product of lengths, find similar triangles. Write the equal ratio first, with the lengths of one triangle on top and the matching lengths of the other below, then cross-multiply. The product $AD \times BC = AB \times DE$ is the same fact as the ratio $\dfrac{AD}{AB} = \dfrac{DE}{BC}$ .

Example 2.4

Similar triangles in a different position

Triangles $A B C$ and ${D} E F$ are drawn apart. $\angle B A C = \angle E {D} F = 50^\circ$ and $\angle A B C = \angle {D} E F = 70^\circ$ . Prove that the triangles are similar, and find $\angle {D} F E$ .

Show solutionHide solution

Colour keythe angle or statement you were asked for

Given:

\angle B A C = \angle E {D} F

Given:

\angle A B C = \angle {D} E F

\triangle A B C \sim \triangle {D} E F

(AA: two pairs of equal angles)

A \to {D}

B \to E

C \to F

: write the matching letters in the same order.

\angle A C B + \angle B A C + \angle A B C = 180^\circ

(angle sum of a triangle). So

\angle A C B

is:

\displaystyle 180^\circ - 50^\circ - 70^\circ = \textcolor{#010106}{60^\circ}

Subtract the known angles from

180^\circ

\angle A C B = \angle {D} F E

(corresponding parts of similar triangles), so

\angle {D} F E = 60^\circ

F

matches

C

Answer

\triangle A B C \sim \triangle {D} E F\ \text{(shown)}

;

\angle {D} F E = 60^\circ

Your Turn 2.4

Triangles $A B C$ and ${D} E F$ are drawn apart. $\angle B A C = \angle E {D} F = 55^\circ$ and $\angle A B C = \angle {D} E F = 60^\circ$ . Prove that the triangles are similar, and find $\angle {D} F E$ .

Show full solutionHide solution

Colour keythe angle or statement you were asked for

Given:

\angle B A C = \angle E {D} F

Given:

\angle A B C = \angle {D} E F

\triangle A B C \sim \triangle {D} E F

(AA: two pairs of equal angles)

A \to {D}

B \to E

C \to F

: write the matching letters in the same order.

\angle A C B + \angle B A C + \angle A B C = 180^\circ

(angle sum of a triangle). So

\angle A C B

is:

\displaystyle 180^\circ - 55^\circ - 60^\circ = \textcolor{#010106}{65^\circ}

Subtract the known angles from

180^\circ

\angle A C B = \angle {D} F E

(corresponding parts of similar triangles), so

\angle {D} F E = 65^\circ

F

matches

C

Answer

\triangle A B C \sim \triangle {D} E F\ \text{(shown)}

;

\angle {D} F E = 65^\circ

Lengths scale by the ratio, areas by its square. If triangle ${D} E F$ is similar to triangle $A B C$ with every length multiplied by $k$ , then the area of ${D} E F$ is $k^2$ times the area of $A B C$ . A ratio of $\dfrac{3}{2}$ between lengths gives a ratio of $\dfrac{9}{4}$ between areas. You use this fact to state a result, not inside a proof of similarity.

Example 2.5

Similar triangles from a trapezium

The straight lines $AB$ and $CD$ cross at $X$ , and $AC$ is parallel to $BD$ , so $A C B {D}$ is a trapezium whose diagonals are $AB$ and $CD$ . The angle $\angle A X C = 80^\circ$ . Prove that triangles $X A C$ and $X B {D}$ are similar and that $XA \times XD = XB \times XC$ , then find $\angle C X B$ .

Show solutionHide solution

Colour keythe angle or statement you were asked for

Given:

AC \parallel BD

\angle C A X = \angle {D} B X

(alternate angles, parallel lines)

AC \parallel BD

with transversal

AB

: alternate angles.

\angle B X {D} = \angle A X C

(vertically opposite angles are equal), so

\angle B X {D} = 80^\circ

The lines

AB

and

CD

cross at

X

\triangle X A C \sim \triangle X B {D}

(AA: two pairs of equal angles)

X \to X

A \to B

C \to {D}

\dfrac{XA}{XB} = \dfrac{XC}{XD}

(corresponding sides of similar triangles), so

XA \times XD = XB \times XC

Cross-multiply the equal ratios.

\angle A X C + \angle C X B = 180^\circ

(angles on a straight line add up to

180^\circ

). So

\angle C X B

is:

\displaystyle 180^\circ - 80^\circ = \textcolor{#010106}{100^\circ}

A

X

B

are on a straight line.

Answer

\triangle X A C \sim \triangle X B {D}\text{ and }XA \times XD = XB \times XC\ \text{(shown)}

;

\angle C X B = 100^\circ

Your Turn 2.5

The straight lines $AB$ and $CD$ cross at $X$ , and $AC$ is parallel to $BD$ , so $A C B {D}$ is a trapezium whose diagonals are $AB$ and $CD$ . The angle $\angle A X C = 60^\circ$ . Prove that triangles $X A C$ and $X B {D}$ are similar and that $XA \times XD = XB \times XC$ , then find $\angle C X B$ .

Show full solutionHide solution

Colour keythe angle or statement you were asked for

Given:

AC \parallel BD

\angle C A X = \angle {D} B X

(alternate angles, parallel lines)

AC \parallel BD

with transversal

AB

: alternate angles.

\angle B X {D} = \angle A X C

(vertically opposite angles are equal), so

\angle B X {D} = 60^\circ

The lines

AB

and

CD

cross at

X

\triangle X A C \sim \triangle X B {D}

(AA: two pairs of equal angles)

X \to X

A \to B

C \to {D}

\dfrac{XA}{XB} = \dfrac{XC}{XD}

(corresponding sides of similar triangles), so

XA \times XD = XB \times XC

Cross-multiply the equal ratios.

\angle A X C + \angle C X B = 180^\circ

(angles on a straight line add up to

180^\circ

). So

\angle C X B

is:

\displaystyle 180^\circ - 60^\circ = \textcolor{#010106}{120^\circ}

A

X

B

are on a straight line.

Answer

\triangle X A C \sim \triangle X B {D}\text{ and }XA \times XD = XB \times XC\ \text{(shown)}

;

\angle C X B = 120^\circ

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Notes version 1.0 · Last updated 3 Nov 2026 · Chapter 17 of 17