Question

In a triangle ABC, BD and CE are perpendiculars to the sides AC and AB respectively. If BD and CE intersect at I, then, $\angle BIC={180}^{\circ }-\angle A$. State true or false

• A. True
• B. False

Answer 1

The correct option is A True

$BD\perp AC$
Hence, $\angle BDA={90}^o$
$CE\perp AB$
Hence,$\angle CEA={90}^o$
In quadrilateral, AEID
$\angle A+\angle AEI+\angle AIE+\angle IDA={360}^o$
$\angle A+90+\angle EID+90={360}^o$
$\angle A+\angle EID={180}^o$
$\angle EID=\angle BIC$ (Vertically opposite angles)
$\angle A+\angle BIC={180}^o$
$\angle BIC={180}^o-\angle A$