In triangle ABC, AB = AC. Point D lies on AB such that

Solution

ATQ, Let BC = 1. Given: AB = AC and AD = DC = BC = 1. Let AB = AC = x, so BD = AB − AD = x − 1. Let ∠A = ∠BAC = α. From ΔABC: cos α = (AB² + AC² − BC²) / (2·AB·AC) cos α = (x² + x² − 1) / (2x²) = (2x² − 1) / (2x²) From ΔADC (AD = DC = 1, AC = x): DC² = AC² + AD² − 2·AC·AD·cos α 1 = x² + 1 − 2x cos α => 2x cos α = x² => cos α = x/2 Equate: x/2 = (2x² − 1)/(2x²) => x³ − 2x² + 1 = 0 => (x − 1)(x² − x − 1) = 0 => x = (1 + √5)/2 (since x > 1) So BD = x − 1 = (√5 − 1)/2 Now in ΔBDC: BC = 1, DC = 1, BD = (√5 − 1)/2 Use cosine rule at D (opposite BC): BC² = BD² + DC² − 2·BD·DC·cos ∠BDC 1 = (x−1)² + 1 − 2(x−1) cos ∠BDC => cos ∠BDC = (x−1)/2 = ((√5 − 1)/2)/2 = (√5 − 1)/4 Since cos 72° = (√5 − 1)/4, ∠BDC = 72°