Question

In Δ ABC, BD ⊥ AC and CE ⊥ AB. If BD and CE intersect at O, prove that ∠BOC = 180° - A.

Answer 1

In Δ ABC, BD ⊥ AC and CE ⊥ AB. If BD and CE intersect at O, we need to prove that ∠BOC = 180° - A.

• Draw a figure according to the instructions in the question.
• In ΔDBC, ∠DBC + ∠C + 90° = 180°
• ∠DBC=∠OBC= 90° - ∠C    (1)
• Similarly, in ΔBEC, 90° + ∠ECB + ∠B = 180°
• ∠ECB=∠OCB= 90°-∠B          (2)      
• In ΔABC, ∠A+∠B+∠C = 180°
• ∠B + ∠C = 180°-∠A                (3)
• In ΔOBC, ∠BOC + ∠OBC + ∠OCB=180°
• From 1 and 2, ∠BOC + 90°-∠C + 90° - ∠B =180°
• ∠BOC = ∠B + ∠C
• From 3, ∠BOC = 180° - ∠A
• hence, proved.

Answer 2

Answer:

ok it's your answer. bye. me