[" pumple "8" : In Fig."6.38" ,the sides "AB" and "AC" of "],[" Af bisectors "BO" and "D" respectively."],[" sespectively meet at point "0" ,then prove that "]

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. , sides AB and AC of DeltaABC are extended to points P and Q respectively. Also anglePBC AB .

In Fig. , sides AB and AC of DeltaABC are extended to points P and Q respectively. Also anglePBC AB .

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 .