Question

In triangle ABC, AD is perpendicular to side BC and AD² = BD × DC. Show that angle BAC = 90°.

In ΔABC, if AD⊥ BC and AD² = BD × DC, then prove that ∠BAC = 90°.

In \(\triangle \mathrm{ABC}, \mathrm{AD}\) is perpendicular to side BC and \(\mathrm{AD}^{2}=\mathrm{BD} \times \mathrm{CD}\), prove that \(\angle \mathrm{BAC}=90^{\circ}\).

Answer 1

Given: AD² = BD × DC

`(AD)/(DC) = (BD)/(AD)`

∠ADB = ∠ADC = 90°

∴ ∆DBA ~ ∆DAC  ...(SAS similarity)

So, these two triangles will be equiangular.

∴ ∠1 = ∠C and ∠2 = ∠B

∠1 + ∠2 = ∠B + ∠C

∠A = ∠B + ∠C    ...(i)

By angle sum property,

∠A + ∠B + ∠C = 180°

∠A + ∠A = 180°  ...[From (i) we get]

2∠A = 180°

∠A = ∠BAC = 90°