# 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

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