Question

in triangle ABC,BD is perpendicular to AC,CE is perpendicular to AB,if BD and CE intersect at O ,prove that angle BOC =180 minus angle A

Answer 1

to prove : BOC= 180-angle A

Proof: In triangle BCE

B+BCE+CEB=180

B+BCE+90=180

BCE= BCO =180-(B+90)

=180-B-90............1

similarly in triangle BCD

DBC=OBC =180-(C+90)

=180-C-90............2

and in triangle ABC

B=180-(A+C)

=180-A-C...........3

now in triangle BOC

BOC + OBC+ BCO =180

BOC=180-(OBC+BCO)

BOC=180-(180-C-90+180-B-90) from 1 and 2

=180-(360-180-C-B)

=180-(180-C-(180-A-C))

=180-(180-C-180+A+C)

=180-180+180-A+C-C

BOC=180-A proved