Triangles

Two figures are said to be similar if they have exactly the same shape but not necessarily the same size. All squares are similar, all circles are similar, and all equilateral triangles are similar. In contrast, congruent figures have the same shape and the same size, so every congruent pair is also similar, but similar figures need not be congruent.

For triangles, similarity has a precise meaning. Two triangles are similar if both of these hold:

• their corresponding angles are equal, and
• their corresponding sides are in the same ratio (proportional).

If △ABC is similar to △DEF we write △ABC ~ △DEF. The order of letters matters: it tells us A↔D, B↔E, C↔F. So ∠A = ∠D, ∠B = ∠E, ∠C = ∠F and AB/DE = BC/EF = CA/FD.

Worked example. Suppose △ABC ~ △PQR with AB = 4 cm, BC = 6 cm, CA = 5 cm and PQ = 8 cm. Since the correspondence is A↔P, B↔Q, C↔R, the ratio of the first pair of sides is AB/PQ = 4/8 = 1/2. Every pair of corresponding sides must share this ratio, so BC/QR = 1/2 gives QR = 12 cm and CA/RP = 1/2 gives RP = 10 cm. The corresponding angles of the two triangles remain equal regardless of the enlargement.

The Basic Proportionality Theorem (BPT), also called Thales' Theorem, states: If a line is drawn parallel to one side of a triangle intersecting the other two sides at distinct points, then it divides those two sides in the same ratio.

In △ABC, if a line DE meets AB at D and AC at E with DE ∥ BC, then AD/DB = AE/EC.

Proof idea (using areas). Join BE and CD, and drop perpendiculars EM ⊥ AB and DN ⊥ AC. Then ar(ADE) = ½ × AD × EM and ar(BDE) = ½ × DB × EM, so ar(ADE)/ar(BDE) = AD/DB. Similarly ar(ADE)/ar(CDE) = AE/EC. But △BDE and △CDE lie on the same base DE and between the same parallels DE and BC, so ar(BDE) = ar(CDE). Comparing the two ratios gives AD/DB = AE/EC.

The Converse of BPT says: If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side. That is, in △ABC, if D lies on AB and E on AC with AD/DB = AE/EC, then DE ∥ BC. This converse is the standard tool for proving two lines are parallel.

Worked example. In △ABC, D and E lie on AB and AC with DE ∥ BC. Given AD = 2 cm, DB = 3 cm and AE = 4 cm, find EC. By BPT, AD/DB = AE/EC, so 2/3 = 4/EC. Cross-multiplying, 2 × EC = 3 × 4 = 12, hence EC = 6 cm.

To prove two triangles similar we do not need to verify all three angles and all three side ratios. Any one of the following criteria is enough.

1. AAA (and AA) similarity. If in two triangles the corresponding angles are equal, the triangles are similar and their corresponding sides are automatically proportional. Because the angles of a triangle add to 180°, knowing two pairs of equal angles forces the third pair to be equal too. So AA is sufficient: if ∠A = ∠D and ∠B = ∠E, then △ABC ~ △DEF.

2. SSS similarity. If the corresponding sides of two triangles are in the same ratio, then their corresponding angles are equal and the triangles are similar: AB/DE = BC/EF = CA/FD ⇒ △ABC ~ △DEF.

3. SAS similarity. If one angle of a triangle equals one angle of the other and the sides including these angles are in the same ratio, the triangles are similar: ∠A = ∠D and AB/DE = AC/DF ⇒ △ABC ~ △DEF. The equal angle must be the angle between the two proportional sides.

Note that 'ASA' and 'SSA' are congruence-style labels and are not separate similarity criteria; AA already covers the angle case.

Worked example (AA). Two right triangles △ABC and △PQR are right-angled at B and Q, and ∠A = ∠P. Then two angles of the first triangle equal two angles of the second (∠B = ∠Q = 90° and ∠A = ∠P), so by the AA criterion △ABC ~ △PQR, and therefore AB/PQ = BC/QR = CA/PR.

When △ABC ~ △DEF, all three pairs of corresponding sides carry the same ratio. This common value is the scale factor (or ratio of similarity), often written k: AB/DE = BC/EF = CA/FD = k.

If k = 1 the triangles are congruent; if k > 1 the first triangle is an enlargement of the second; if k < 1 it is a reduction. Corresponding sides are the sides opposite equal angles — reading the similarity statement in order tells you which side matches which.

The scale factor governs more than just the sides. The ratio of the perimeters of two similar triangles equals the ratio of their corresponding sides, and so does the ratio of any pair of corresponding line segments such as medians, altitudes or angle bisectors. This is because each such segment scales up or down by the same factor k as the triangle itself.

Worked example. △ABC ~ △DEF with AB = 5 cm, BC = 7 cm, CA = 6 cm and DE = 10 cm. The scale factor is AB/DE = 5/10 = 1/2, meaning △ABC is half the size of △DEF. Hence EF = BC × 2 = 14 cm and FD = CA × 2 = 12 cm. Check the perimeters: perimeter of ABC = 5 + 7 + 6 = 18 cm and perimeter of DEF = 10 + 14 + 12 = 36 cm, giving 18/36 = 1/2, exactly the scale factor.

The main practical use of similarity is finding unknown lengths. Once two triangles are known to be similar, every pair of corresponding sides is in the same ratio, so a single proportion solves for the missing side. The method has three steps: (1) establish similarity (by AA, SSS, SAS, or by BPT for a parallel line), (2) write the correct ratio matching corresponding sides, and (3) cross-multiply to solve.

Heights and shadows. A tall object and its shadow, together with the Sun's rays, form a triangle similar to that of a short object standing nearby, because the Sun's rays strike both at the same angle. So (height of tall object)/(its shadow) = (height of short object)/(its shadow).

Worked example 1 (shadow). A vertical pole 6 m high casts a shadow 4 m long on the ground, and at the same time a tower casts a shadow 28 m long. Let h be the height of the tower. The two triangles are similar (equal Sun angle and both upright), so 6/4 = h/28. Then h = (6 × 28)/4 = 168/4 = 42 m.

Worked example 2 (parallel line). In △ABC, DE ∥ BC with D on AB and E on AC. Given AD = 3 cm, AB = 9 cm and AC = 12 cm, find AE. Using the whole-side form of BPT, AD/AB = AE/AC, so 3/9 = AE/12. Thus AE = (3 × 12)/9 = 36/9 = 4 cm.

Worked example 3 (similar triangles). △ABC ~ △PQR with AB = 6 cm, BC = 9 cm and PQ = 4 cm. Find QR. Matching corresponding sides, AB/PQ = BC/QR, so 6/4 = 9/QR, giving QR = (9 × 4)/6 = 36/6 = 6 cm.