Question

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

Answer 1

Given,

BD perpendicular to AC

And,

CE perpendicular to AB

In ΔBCE

∠E + ∠B + ∠ECB = 180°

90° + ∠B + ∠ECB = 180°

∠B + ∠ECB = 90°

∠B = 90° - ∠ECB .....(i)

In ΔBCD

∠D + ∠C + ∠DBC = 180°

90° + ∠C + ∠DBC = 180°

∠C + ∠DBC = 90°

∠C = 90° - ∠DBC .....(ii)

Adding (i) and (ii), we get

∠B + ∠C = 180°(∠ECB + ∠DBC)

∠180° - ∠A = 180°(∠ECB + ∠DBC)

∠A = ∠ECB + ∠DBC

∠A = ∠OCB + ∠OBC (Therefore, ∠ECB = ∠OCB and ∠DCB = ∠OCB) .... (iii)

In ΔBOC

∠BOC + (∠OBC + ∠OCB) = 180°

∠BOC + ∠A = 180°[From (iii)]

∠BOC = 180° - ∠A

Hence, proved