Criteria for similarity of triangles

Checking similarity by definition , equal angles AND proportional sides , is wasteful. Three concise criteria let us prove similarity from much less.

The three criteria

AAA (or AA) Criterion. If two triangles have all three pairs of corresponding angles equal, then they are similar.

Because the angles of a triangle sum to $180^\circ$, knowing two pairs of equal angles automatically gives the third. So this criterion is usually called the AA criterion: two equal angles is enough.

SSS Criterion. If the three pairs of corresponding sides of two triangles are in the same ratio, the triangles are similar.

SAS Criterion. If one pair of corresponding angles is equal and the two sides including those angles are in the same ratio, the triangles are similar.

Outline of proofs

The AA criterion follows from constructing on one triangle a copy of the other angle and using BPT.

Sketch. Let $\triangle ABC$ and $\triangle DEF$ satisfy $\angle A = \angle D$ and $\angle B = \angle E$. Construct on $DE$ a point $P$ with $DP = AB$ and on $DF$ a point $Q$ with $DQ = AC$. Then $\triangle DPQ \cong \triangle ABC$ by SAS. The triangle $\triangle DPQ$ sits inside $\triangle DEF$ with $\angle DPQ = \angle B = \angle E$. By the converse of corresponding angles (parallels), $PQ \parallel EF$. By BPT, $DP/PE = DQ/QF$. Combine to get all sides proportional. $\blacksquare$ (Sketch.)

The SSS and SAS criteria are proved similarly, by an explicit construction that reduces them to the AA case.

Using the criteria

To prove $\triangle ABC \sim \triangle PQR$:

• AA: find two angles of one equal to two angles of the other.
• SSS: show $AB/PQ = BC/QR = CA/RP$.
• SAS: show one angle equal AND the two enclosing sides proportional.

For board questions, AA is the workhorse because angles are often given or easily computed. SSS shows up when only side lengths are given. SAS handles "one common angle, two sides" set-ups.

Consequences

• The medians, altitudes, angle bisectors, and perpendicular bisectors of similar triangles are themselves in the same ratio as the corresponding sides.
• The angle bisector theorem: in $\triangle ABC$, if $AD$ bisects $\angle A$ and $D$ is on $BC$, then $BD/DC = AB/AC$. (Proved via parallels + BPT.)
• A right triangle, with the altitude dropped from the right angle, gets split into two smaller triangles each similar to the original , the foundation of the Pythagorean proof.

Worked examples

Example 1. In $\triangle ABC$ and $\triangle DEF$, $\angle A = \angle D = 80^\circ$ and $\angle B = \angle E = 60^\circ$. Are they similar?

By AA: yes. Two pairs of equal angles automatically give the third.

Example 2. Two triangles have sides $4, 6, 8$ and $6, 9, 12$. Are they similar?

Compute ratios: $4/6 = 2/3$, $6/9 = 2/3$, $8/12 = 2/3$. All equal , SSS, similar.

Example 3. In $\triangle ABC$, $AB = 6, AC = 8$. In $\triangle DEF$, $DE = 9, DF = 12$. $\angle A = \angle D$. Are they similar?

$AB/DE = 6/9 = 2/3$ and $AC/DF = 8/12 = 2/3$. With $\angle A = \angle D$ , SAS, similar.

Example 4. A vertical pole of height $6$ m casts a shadow of $4$ m. At the same time a tower casts a shadow $28$ m long. Find the height of the tower.

Sun's angle is the same. Two right triangles share an angle $\Rightarrow$ AA $\Rightarrow$ similar. Ratio of heights to shadows is constant.

$\dfrac{\text{tower height}}{28} = \dfrac{6}{4} \Rightarrow$ tower height $= 42$ m.

Example 5. In $\triangle ABC$, $\angle ACB = 90^\circ$ and $CD \perp AB$ with $D$ on $AB$. Prove $\triangle ACB \sim \triangle ADC$ and $\triangle ACB \sim \triangle CDB$.

In $\triangle ACB$ and $\triangle ADC$: $\angle ACB = \angle ADC = 90^\circ$ and $\angle A$ common. By AA, $\triangle ACB \sim \triangle ADC$.

Similarly, $\angle ACB = \angle BDC = 90^\circ$ and $\angle B$ common $\Rightarrow$ $\triangle ACB \sim \triangle CDB$ by AA. ✓

Try it yourself

1. State the three criteria for triangle similarity.
2. In $\triangle ABC$ and $\triangle XYZ$, $\angle B = \angle Y = 70^\circ$, $AB/XY = BC/YZ$. Justify $\triangle ABC \sim \triangle XYZ$.
3. Show two triangles with sides $3, 4, 5$ and $9, 12, 15$ are similar.
4. In $\triangle ABC$, $D$ on $AC$ such that $\angle ADB = \angle C$. Prove $\triangle ABD \sim \triangle ACB$.
5. The shadow of a $6$-m pole is $4$ m. A man $1.8$ m tall casts a shadow $1.2$ m. Are the sun's elevation triangles similar? Verify.
6. In $\triangle PQR$, $S$ on $PQ$, $T$ on $PR$, $PS \cdot PR = PQ \cdot PT$. Prove $\triangle PST \sim \triangle PQR$.
7. State and prove the angle bisector theorem using AA.
8. Two right triangles, one with legs $3, 4$, another with legs $9, 12$. Similar?
9. In $\triangle ABC$, $AD$ is the altitude on $BC$. Prove $\triangle BDA \sim \triangle ADC$ (when $\angle A = 90^\circ$).
10. A stick of length $1$ m casts a shadow of $40$ cm. At the same time a tower's shadow is $30$ m. Find the tower's height.

Pitfalls / Insight

• Match the vertices when writing the similarity statement. $\triangle ABC \sim \triangle XYZ$ means $A \leftrightarrow X, B \leftrightarrow Y, C \leftrightarrow Z$.
• AAA needs all three equal angles, but AA is enough because angles sum to $180^\circ$.
• SAS needs the angle between the proportional sides , the included angle, not just any pair.

Insight. AA is the easiest, fastest similarity test in board exams. Look for a common angle and a parallel line, and you're nearly done.