Question

In $\Delta ABC,BD\perp AC$ and CE $\perp AB.$ If BD and CE intersect at O , prove that $\angle BOC={180}^{\circ }-\angle A.$

Answer 1

Given : In $\Delta ABC,BD\perp AC$ and $\perp ABBD$ and CE interesect each other at O

To prove : $\angle BOC={180}^{\circ }-\angle A$

Proof : In quadrilateral ADOE

$\angle A+\angle D+\angle DOE+\angle E={360}^{\circ }$

(Sum of angles of quadrilateral)

$\Rightarrow \angle A+{90}^{\circ }+\angle DOE+{90}^{\circ }={360}^{\circ }$

$\Rightarrow \angle A+\angle DOE={360}^{\circ }-{90}^{\circ }-{90}^{\circ }={180}^{\circ }$

But $\angle BOC=\angle DOE$

(Vertically opposite angles)

$\Rightarrow \angle A+\angle BOC={180}^{\circ }$

$\therefore \angle BOC={180}^{\circ }-\angle A$