In $\triangle ABC, BD \perp AC$ and $CE \perp AB$. If $BD$ and $CE$ intersect at $O$, prove that $\angle BOC = 180^o-\angle A$.
Given:
In $\triangle ABC, BD \perp AC$ and $CE \perp AB$.
$BD$ and $CE$ intersect at $O$.
To do:
We have to prove that $\angle BOC = 180^o-\angle A$.
Solution:
From the figure,
In quadrilateral $ADOE$,
$\angle A + \angle D + \angle O + \angle E = 360° (Sum of angles of quadrilateral)
$\angle A + 90^o + \angle DOE + 90^o = 360^o$
$\angle A + \angle DOE = 360^o - 180^o = 180^o$
$\angle BOC = \angle DOE$ (Vertically opposite angles)
This implies,
$\angle A + \angle BOC = 180^o$
$\angle BOC = 180^o - \angle A$
Hence proved.
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