Question:
In triangle ABC, BD ⊥ AC and CE ⊥ AB. If BD and CE intersect at O, prove that ∠BOC = 180° - ∠A.
Solution:
In quadrilateral AEOD
∠A + ∠AEO + ∠EOD + ∠ADO = 360°
⇒ ∠A + 90° + 90° + ∠EOD = 360°
⇒ ∠A + ∠BOC = 180° [∵ ∠EOD = ∠BOC vertically opposite angles]
⇒ ∠BOC = 180° − ∠A
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