Proving Triangles Are Similar

Prove two triangles are similar by finding two pairs of equal angles and giving the reason for each.

In the exam for Cambridge IGCSE Maths 0580, Cambridge O Level Maths 4024 · 6 real past-paper questions below

Two Angles Are Enough

To prove two triangles are similar, show that two pairs of angles are equal. The third pair then matches too, because each triangle's angles add up to 180°.

Every equal pair needs a reason. These are the four you will use most:

Reason	Where you see it
common angle	one corner in both triangles
corresponding angles	parallel lines, F shape
alternate angles	parallel lines, Z shape
vertically opposite angles	two lines crossing

Rule

Write three lines: first pair and reason, second pair and reason, then "so the triangles are similar".

The A Shape Proof

The 3-mark question: $D E$ is parallel to $B C$ , with $D$ on $A B$ and $E$ on $A C$ . Prove triangles $A D E$ and $A B C$ are similar.

Worked example

1
$∠ D A E = ∠ B A C$ — Both triangles contain the corner $A$ , so this angle is in each one. It is a common angle.
2
$∠ A D E = ∠ A B C$ — $D E$ is parallel to $B C$ , and these angles sit in the same place on each line, an F shape along $A B$ . So they are corresponding angles, and equal.
3
$so A D E and A B C are similar$ — Two pairs of equal angles is enough. The third pair must match too, because the angles in a triangle add up to 180°.

Common mistake

Saying "alternate angles" here. Angles $A D E$ and $A B C$ sit on the same side of the parallel lines in an F shape, so they are corresponding.

Key idea

Each pair of equal angles needs its own reason, and each correct pair with its reason earns a mark. Say “so the triangles are similar” only after you have two pairs.

The X Shape Proof

The 3-mark question: lines $A E$ and $B D$ cross at $C$ , and $A B$ is parallel to $D E$ . Prove triangles $A B C$ and $E D C$ are similar.

Worked example

1
$∠ A C B = ∠ E C D$ — The lines $A E$ and $B D$ cross at $C$ , so the angles facing each other there are vertically opposite angles, and equal.
2
$∠ B A C = ∠ D E C$ — $A B$ is parallel to $D E$ , and these angles sit in a Z shape between the parallel lines, on opposite sides of $A E$ . So they are alternate angles, and equal.
3
$so A B C and E D C are similar$ — Two pairs of equal angles prove it. Write the letters in matching order: $A$ with $E$ , $B$ with $D$ , $C$ with $C$ .

Common mistake

Naming the second triangle in the wrong order. $A$ matches $E$ , so write $E D C$ , not $D E C$ . The order tells the examiner which sides match.

A Shared Angle, No Parallel Lines

The harder question gives one pair of equal angles and hides the other. $D$ is on $A C$ , and angle $A B D$ equals angle $A C B$ . We are told $A B = 6$ cm and $A C = 9$ cm, and we want $A D$ .

Worked example

1
$∠ B A D = ∠ C A B$ — $D$ is on $A C$ , so the angle at $A$ is the same in triangles $A B D$ and $A C B$ . It is a common angle.
2
$∠ A B D = ∠ A C B$ — The question tells us this, so the reason is “given”.
3
$so A B D and A C B are similar$ — Two pairs of equal angles. The letters match as $A$ with $A$ , $B$ with $C$ and $D$ with $B$ .
4
$\frac{A D}{A B} = \frac{A B}{A C}$ — Matching sides of similar triangles are in the same ratio. $A D$ matches $A B$ , and $A B$ matches $A C$ .
5
$\frac{A D}{6} = \frac{6}{9}$ — Put in $A B = 6$ and $A C = 9$ .
6
$A D = \frac{6 \times 6}{9}$ — Multiply both sides by 6 to get $A D$ on its own.
7
$A D = 4 cm$ — 36 ÷ 9 = 4.

Key idea

Now use the scale factor to find lengths: see Similar Triangles.

Other Places to Find Equal Angles

Some proofs have points on a circle, or a rhombus. You still need two equal pairs of angles, but the reasons come from other facts.

Reason	Where you see it
angles on a straight line	add up to 180°
opposite angles of a cyclic quadrilateral	add up to 180°, they are not equal
angles in the same segment	two angles from the same chord, on the same side
opposite sides of a rhombus or parallelogram	parallel, so look for alternate angles

$P$ , $Q$ , $R$ , $S$ lie on a circle in that order. The lines $P Q$ and $S R$ meet at $T$ outside the circle. Show that triangle $T QR$ is similar to triangle $T S P$ .

Worked example

1
$∠ QT R = ∠ S T P$ — Angle $T$ is in both triangles, so it is a common angle.
2
$∠ T QR + ∠ P QR = 18 0^{\circ}$ — $P$ , $Q$ and $T$ are on a straight line, so these angles on a straight line add up to 180°.
3
$∠ P QR + ∠ P S R = 18 0^{\circ}$ — $P QR S$ is a cyclic quadrilateral, so its opposite angles add up to 180°.
4
$∠ T QR = ∠ P S R$ — Both add to 180° with the same angle $P QR$ , so they are equal.
5
$∠ P S R = ∠ T S P$ — $R$ is on the line $S T$ , so these are the same angle.
6
$so T QR and T S P are similar$ — Two pairs of equal angles. $Q$ matches $S$ and $R$ matches $P$ .

Common mistake

Saying opposite angles of a cyclic quadrilateral are equal. They add up to 180°.

Which Method? Read the Question

Look for parallel lines and crossing lines first.

You see	Reasons to use
A line parallel to a side	common angle, corresponding angles
Two lines crossing	vertically opposite, alternate angles
One equal pair given	given, plus common angle
Points on a circle	straight line, cyclic quadrilateral, same segment
Then find a length	Similar Triangles
Prove congruent, not similar	Congruent Shapes

Key idea

Give the reason in words the examiner knows. "Z angles" and "F angles" do not score.

Practise proving triangles are similar now

Interactive visual aids, then real past-paper questions marked on your working like the exam. Free to start.

Practise free on Lemmara

Real past-paper questions on proving triangles are similar

Each with a full worked solution.

Lemmara is an independent study tool, not affiliated with or endorsed by Cambridge International Education or Pearson Edexcel. The lessons and worked examples are Lemmara’s own.