Question
In triangle $ABC, AD$ is perpendicular to side $BC$ and $AD^2 = BD \times DC$.
Show that angle $BAC = 90°$
Show that angle $BAC = 90°$
✓
Answer
Given $AD^2 = BD \times DC$
$\frac{A D}{D C}=\frac{B D}{A D}$
∠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°
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