In triangle ABC, AD is perpendicular to side BC and AD 2 = BD × CD. Prove that angle BAC = 90°.

Triangle ABC is shown in the figure below: Given : AD2 = BD × CD ⇒ ∠ADB = ∠ADC [Both = 90°] ∴ △DBA ~ △DAC (By SAS). Since, triangles are similar they will be equiangular. ∴ ∠1 = ∠C and ∠2 = ∠B ⇒ ∠1 + ∠2 = ∠B + ∠C ⇒ ∠A = ∠B + ∠C By angle sum property : ⇒ ∠A + ∠B + ∠C = 180° ⇒ ∠A + ∠A = 180° ⇒ 2∠A = 180° ⇒ ∠A = 90°. From figure, ⇒ ∠BAC = ∠A = 90°. Hence, proved that ∠BAC = 90°.