Problem solving: Constructing triangles Exercise 5C 1 PS is common RS = PQ (given) RSP = QPS (alternate angles) So (SAS) ∆PRS ≡ ∆PQS 2 BD is common AB = BC (given) BAD = BCD = 90° So (RHS) ∆ABD ≡ ∆BCD 3 Consider and ∆XYP ∆XZP. XP is common to both triangles XY= XZ (given) PY = PZ (given) So (SSS) ∆XYP ≡ ∆XZP Hence XYP = XZP 4 Consider and ∆ABD ∆BCD. BD is common to both triangles AB = BC (given) AD = CD (given) So (SSS) ∆ABD ≡ ∆BCD Hence A = C 5 AB = CD (given)
Problem solving: Constructing triangles Exercise 5C 1 PS is common RS = PQ (given) RSP = QPS (alternate angles) So ∆PRS ≡ ∆PQS (SAS) 2 BD is common AB = BC (given) BAD = BCD = 90° So ∆ABD ≡ ∆BCD (RHS) 3 Consider ∆XYP and ∆XZP. XP is common to both triangles XY= XZ (given) PY = PZ (given) So ∆XYP ≡ ∆XZP (SSS) Hence XYP = XZP 4 Consider ∆ABD and ∆BCD. BD is common to both triangles AB = BC (given) AD = CD (given) So ∆ABD ≡ ∆BCD (SSS) Hence A = C 5 AB = CD (given)