Triangles and Similarity
The exam tests the criteria for similar triangles, the Basic Proportionality Theorem (Thales' theorem), and the relationship between the areas of similar triangles.
Core rules
- AA similarity: two equal corresponding angles are enough. SSS and SAS similarity compare corresponding sides and the included angle.
- Basic Proportionality Theorem: in triangle ABC, if DE ∥ BC with D on AB and E on AC, then AD/DB = AE/EC.
- Converse: if AD/DB = AE/EC, then DE ∥ BC.
- Areas of similar triangles are in the ratio of the squares of corresponding sides.
- Corresponding medians, altitudes and perimeters of similar triangles are in the ratio of corresponding sides.
Worked example
In triangle ABC, DE ∥ BC. AD = 3 cm, DB = 5 cm and AE = 4.5 cm. Find EC. By the Basic Proportionality Theorem, AD/DB = AE/EC, so 3/5 = 4.5/EC. Cross-multiply: 3 × EC = 22.5, hence EC = 7.5 cm. The full side AC is AE + EC = 12 cm, and AB = 8 cm. The ratio of the triangles' areas is (AD/AB)² = (3/8)² = 9/64.
Traps
- Writing corresponding vertices in the wrong order, which corrupts the ratio.
- Using AD/AB instead of AD/DB in the BPT statement.
- Treating the area ratio as equal to the side ratio rather than its square.
- Assuming DE ∥ BC without justification.