Question

in the given figure, the sides ab and AC of of triangle ABC are produced to points p and D respectively. If the bisector BO and CO of angle CBE and angle BCD respectively meet at point O, then prove that angle BOC = 90°-1/2 angle ​

Answer 1

Solution:-

$\sf{ray \: BO \: is \: the \: bisector \: of \angle \: CBE}$

$\therefore \angle \: CBO \: = \frac{1}{2} \angle \:CBE$

$= \frac{1}{2} (180 \degree - y) \\ \\ = 90 \degree \: - \frac{y}{2} ...........(1)$

Similarly , Ray CO is the bisector of

$\angle \: BCD \: \\ \\ \therefore \: \: \angle \: BCO = \frac{1}{2} (180 \degree - z) \\ \\ = 90 \degree \: - \frac{z}{2}$

$\rm{ \: in \triangle \: BOC \: } \\ \\ \angle \: BOC + \angle \: BCO + \angle \: CBO = 180 \degree$

Substituting (1) and (2) in (3), we get ,

$\angle BOC \: + 90 \degree - \frac{z}{2} + 90 \degree - \frac{y}{2} = 180 \degree$

$\angle \: BOC = \frac{z}{2} + \frac{y}{2} \\ \\ \angle \: BOC = \frac{1}{2} (y + z) \\ \\ x + y + z = 180 \degree \\ \\ y + z = 180 \degree - x \\ \\ \\ \rm{therefore \: (4) \: becomes} \\ \\ \angle \:BOC \: = \frac{1}{2} (180 \degree - x) \\ \\ = 90 \degree - \frac{x}{2} \\ \\ = 90 \degree - \frac{1}{2} \angle \: BAC \:$

Answer 2

Answer:

$90 - \frac{1}{2} \angle \: bac$