Question

in triangle ABC BD is perpendicular to AC and CE is perpendicular to AB . if BD and CE intersect at o, prove that angle BOC=180-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

Answer 2

Answer:

Step-by-step explanation:

ADOE is quadrilateral

As;it has four angles,four vertices nad four sides.

Therefore,

Angle AEO +angle ADO +angle EOD+angle A=360

90(given)+90(given)+angle EOD+angle A=360

180+angle EOD+angle A=360

angle EOD+angle A=180

Angle EOD=180-angle A

Angle EOD=angle BOC=180-angle A