Question

In the adjoining figure, the sides AB and AC of ΔABC are produced to points E and D respectively. If bisectors BO and CO of ∠CBE and ∠BCD respectively meet at point O, then prove that ∠BOC = 900−1/2∠BAC.

Answer 1

Answer:

Thus r = ∠ BOC = 90° 1/2∠ BAC

Step-by-step explanation:

As shown in the figure let

∠OBC=p ,∠OCB =∠BOC=r

Now angle ∠EBC=∠BAC+∠BCA( external angle)

So p=1/2(∠BAC+∠BCA)..................(1)

Also ∠ DCB=∠ BAC +∠ ABC

So q=1/2(∠ BAC +∠ ABC)...................(2)

Now in the Δ OBC

p+q+r=180

or 1/2(∠BAC+∠BCA) +1/2(∠ BAC +∠ ABC)+r=180

1/2( ∠BAC+∠BCA+∠ ABC) +1/2*∠ BAC +r=180

1/2*180 +1/2∠ BAC+r=180

r=180-180/2-1/2∠ BAC

Thus r = ∠ BOC = 90° 1/2∠ BAC