Question

In fig 6.38, the sides AB and AC of ∆ABC are produced to point E and D respectively If bisector BO and CO of angle CBE and angle
BCD respectively meet at point O , then prove that angle BOC=90° - 1/2 angle BAC

Answer 1

HEY MATE HERE IS YOUR ANSWER

Ray BO is the bisector of /_CBE

Therefore, /_CBO= 1/2 /_CBE

= 1/2 (180°-Y)
= 90° - Y/2 (1)


Similarly, ray CO bisector of /_BCD

Therefore, /_ BCO =1/2 /_BCD
/_BCO=1/2 /_BCD
=1/2 (180°-Z)

90°- Z/2----------------(2)

In ⛛ BOC, /_BOC+/_BCO+/_CBO = 180°------(3)

Substitute (1) and (2) in (3), you get

/_BOC + 90 - Z/2 + 90 = 180

/_BOC= Z/2 + Y/2
/_BOC= 1/2 (Y+Z) - - - - - - (4)

x+y+z = 180{A.S.P}
y+z = 180-x

Therefore (4) becomes

/_BOC = 1/2 (180-X)

= 90 - x/2

= 90 - 1/2 /_BAC

HENCE, PROVED

Answer 2

Your answer is in the attachment and the earlier answer have some mistakes