Triangles — chapter notes

Two figures are similar when one is an exact enlargement or reduction of the other: they have the same shape, but not necessarily the same size. Two figures are congruent when they have the same shape and the same size.

Two polygons with the same number of sides are similar if and only if

1. all their corresponding angles are equal, and
2. all their corresponding sides are in the same ratio.

Both conditions matter. A square and a non-square rectangle have equal angles but are not similar; a square and a non-square rhombus have proportional sides but are not similar.

Always similar: any two circles, any two squares, any two equilateral triangles, any two line segments. Not always similar: rectangles, rhombuses, isosceles triangles, right triangles.

Every pair of congruent figures is similar (the ratio is 1), but similar figures need not be congruent.

Not in the 2026–27 syllabus: areas of similar triangles, and the proofs of Pythagoras' theorem and its converse.

$\triangle ABC \sim \triangle PQR$ is read "triangle $ABC$ is similar to triangle $PQR$". The order of the letters gives the correspondence: $A \leftrightarrow P$, $B \leftrightarrow Q$, $C \leftrightarrow R$. Therefore

• $\angle A = \angle P$, $\angle B = \angle Q$, $\angle C = \angle R$, and
• $\dfrac{AB}{PQ} = \dfrac{BC}{QR} = \dfrac{CA}{RP}$.

Side $AB$ joins the 1st and 2nd vertices, so it matches $PQ$, which joins the 1st and 2nd vertices of the other triangle. Never pair letters alphabetically: from $\triangle ABC \sim \triangle QRP$ you get $\dfrac{AB}{QR} = \dfrac{BC}{RP} = \dfrac{CA}{PQ}$.

Because every side is multiplied by the same $k$, the perimeters of similar triangles are also in the ratio $k$.

Theorem. If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.

In $\triangle ABC$ with $D$ on $AB$, $E$ on $AC$ and $DE \parallel BC$: $\dfrac{AD}{DB} = \dfrac{AE}{EC}$.

Proof outline (learn it — it is examined).

• Construction: join $BE$ and $CD$; draw $EM \perp AB$ and $DN \perp AC$.
• $\dfrac{\operatorname{ar}(ADE)}{\operatorname{ar}(BDE)} = \dfrac{AD}{DB}$, because both triangles have height $EM$ … (1)
• $\dfrac{\operatorname{ar}(ADE)}{\operatorname{ar}(DEC)} = \dfrac{AE}{EC}$, because both triangles have height $DN$ … (2)
• $\operatorname{ar}(BDE) = \operatorname{ar}(DEC)$: same base $DE$, between the same parallels $DE$ and $BC$ … (3)
• From (1), (2), (3): $\dfrac{AD}{DB} = \dfrac{AE}{EC}$.

Other forms (add 1 to both sides and invert): $\dfrac{AD}{AB} = \dfrac{AE}{AC}$ and $\dfrac{DB}{AB} = \dfrac{EC}{AC}$. Use the part-to-whole form when a whole side is given.

Caution: BPT is about the two cut sides. It does not give the length of $DE$. For that you need similarity: $\dfrac{DE}{BC} = \dfrac{AD}{AB}$ (section 6).

Consequence: three (or more) parallel lines cut any two transversals in the same ratio. Prove it by joining a point on one transversal to a point on the other and applying BPT twice.

Theorem (statement only). If a line divides any two sides of a triangle in the same ratio, the line is parallel to the third side.

$\dfrac{AD}{DB} = \dfrac{AE}{EC} \Longrightarrow DE \parallel BC$.

How to use it. To show two lines are parallel: (1) find both ratios separately, (2) show that they are equal, (3) conclude "$DE \parallel BC$ (converse of BPT)". If the ratios are unequal, the line is not parallel, because a parallel line would have to give equal ratios.

The proof of the converse is not required in the 2026–27 syllabus.

For triangles, fewer facts are enough. State these criteria exactly; their proofs are not asked.

• AAA / AA. If the corresponding angles of two triangles are equal, the triangles are similar. Since the angles of a triangle add up to $180^\circ$, two pairs of equal angles are enough (AA).
• SSS. If the corresponding sides of two triangles are in the same ratio, the triangles are similar (and their corresponding angles are equal).
• SAS. If one angle of a triangle equals one angle of another and the sides including these angles are in the same ratio, the triangles are similar.

SSA is not a criterion: two sides and a non-included angle can fit two different triangles.

Where equal angles hide in figures: a common angle; vertically opposite angles; corresponding or alternate angles from parallel lines; right angles from perpendiculars, vertical poles or walls.

Method.

1. Name the two triangles and give the criterion with reasons ("$\angle A$ common; $\angle ADE = \angle ABC$, corresponding angles").
2. Write the similarity statement in the correct order.
3. Write all three side ratios from that order.
4. Use the pair of sides you know completely to find $k$, then find the unknown side. Check that all ratios agree.

A parallel line inside a triangle. If $DE \parallel BC$ in $\triangle ABC$, then $\triangle ADE \sim \triangle ABC$ (AA), so $\dfrac{AD}{AB} = \dfrac{AE}{AC} = \dfrac{DE}{BC}$. The parallel segment compares with the whole side.

Crossing lines (X shape). When $AC$ and $BD$ cross at $O$ and $AB \parallel CD$, $\triangle AOB \sim \triangle COD$ (alternate angles and vertically opposite angles). This is how trapezium diagonal problems are solved.

"Crossed" similarity. If $\dfrac{PS}{PR} = \dfrac{PT}{PQ}$ with a common angle $P$, then $\triangle PST \sim \triangle PRQ$ — $S$ matches $R$. Then $ST$ is usually not parallel to $QR$.

Sun and shadows. The sun's rays are parallel, so at one moment every vertical object on level ground forms a right triangle with its shadow, and all these triangles have the same angles (AA). Hence $\dfrac{\text{height}}{\text{shadow}}$ is the same for every object.

Lamp posts. Light spreads from a point. A person of height $h$ standing $d$ from a post of height $H$ has shadow $x$ with $\dfrac{h}{H} = \dfrac{x}{x + d}$, because both triangles share the angle at the shadow tip.

Mirror method. A mirror laid flat on level ground: the angle of incidence equals the angle of reflection, so the eye triangle and the object triangle are similar: $\dfrac{\text{object height}}{\text{object–mirror distance}} = \dfrac{\text{eye height}}{\text{eye–mirror distance}}$.

Sighting over a stick. When the eye is above the ground, the similar triangles start at eye level: subtract the eye height from the vertical lengths first.

Conditions to state: same moment (for shadows), level ground, objects vertical.

Board marking schemes award marks for each justified step, so a proof is graded line by line.

1. Figure — draw and label it, including any construction (usually 1 mark).
2. Given / To prove / Construction — one line each.
3. Statements with reasons — one fact per line; the reason in brackets: (common), (vertically opposite angles), (alternate angles, $AB \parallel DC$), (BPT), (converse of BPT).
4. Criterion — "$\therefore \triangle APR \sim \triangle ABD$ (AA)", with the vertices in matching order.
5. Conclusion — finish with exactly what was asked, then "Hence proved."

A correct final line with no reasons earns little; a well-reasoned proof with a small slip at the end still earns most of the marks.