In Δ ABC, BD ⊥ AC and CE ⊥ AB. If BD and CE intersect at O, prove that ∠BOC = 180° - A.
Given,
BD perpendicular to AC
And,
CE perpendicular to AB
In
∠E + ∠B + ∠ECB = 180o
90o + ∠B + ∠ECB = 180o
∠B + ∠ECB = 90o
∠B = 90o - ∠ECB (i)
In
∠D + ∠C + ∠DBC = 180o
90o + ∠C + ∠DBC = 180o
∠C + ∠DBC = 90o
∠C = 90o - ∠DBC (ii)
Adding (i) and (ii), we get
∠B + ∠C = 180o (∠ECB + ∠DBC)
∠180o - ∠A = 180o (∠ECB + ∠DBC)
∠A = ∠ECB + ∠DBC
∠A = ∠OCB + ∠OBC (Therefore, ∠ECB = ∠OCB and ∠DCB = ∠OCB) (iii)
In
∠BOC + (∠OBC + ∠OCB) = 180o
∠BOC + ∠A = 180o [From (iii)]
∠BOC = 180o - ∠A
Hence, proved