In Fig. 6.38, the sides AB and AC of ABC 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 `/_B O C=90o-1/2/_B A C` .

In Fig. 6.38, the sides AB and AC of ABC 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 /_B O C=90^(@)-1/2/_B A C .

In Fig. 6.38, the sides AB and AC of Delta ABC are produced to points E and D respectively. If bisectors BO and CO of angle CBE and angle BCD respectively meet at point O, then prove that angle BOC = 90^@ -1/2 angle BAC .

In the adjacent figure the sides AB and AC of Delta ABC are produced to points E and D respectively. If bisectors BO and CO of angleCBE and angleBCD respectively meet at point O, then prove that angleBOC = 90^(@)-(1)/(2) angleBAC .

In the adjacent figure the sides AB and AC of Delta ABC are produced to points E and D respectively. If bisectors BO and CO of angleCBE and angleBCD respectively meet at point O, then prove that angleBOC = 90^(@)-(1)/(2) angleBAC .

In the adjacent figure the sides AB and AC of Delta ABC are produced to points E and D respectively. If bisectors BO and CO of angleCBE and angleBCD respectively meet at point O, then prove that angleBOC = 90^(@)-(1)/(2) angleBAC .

In the adjacent figure the sides AB and AC of Delta ABC are produced to points E and D respectively. If bisectors BO and CO of angleCBE and angleBCD respectively meet at point O, then prove that angleBOC = 90^(@)-(1)/(2) angleBAC .

In the given Fig , Sides AB and AC of a DeltaABC are produced to E and D respectively. If respective bisectors BO and CO of angleCBE and angleBCD intersect each other at point O, prove that angleBOC = 90^@-1/2 angleBAC .

In the given Fig , Sides AB and AC of a DeltaABC are produced to E and D respectively. If respective bisectors BO and CO of angleCBE and angleBCD intersect each other at point O, prove that angleBOC = 90^@-1/2 angleBAC .

The sides A B and A C of A B C are product to P and Q\ respectively. the bisectors of exterior angles at B and C of A B C meet at O\ (fig.19) prove that /_B O C=90^0-1/2/_A

The sides AB and AC of ABC are product to P and Q respectively.the bisectors of exterior angles at B and C of ABC meet at O( fig..19) prove that /_BOC=90^(@)-(1)/(2)/_A