Q. 104.1( 19 Votes )

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

Answer :



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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