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

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 .

The sides A B and A C of a A B C are produced to P and Q respectively. If the bisectors of /_P B C AND /_Q C B intersect at O , then /_boc=90^0-1/2/_A GIVEN : A A B C in which sides A B and A C are produced to P and Q respectively. The bisectors of /_P B C and /_Q C B intersect at Odot

The sides A B and A C of a A B C are produced to P and Q respectively. If the bisectors of /_P B C AND /_Q C B intersect at O , then /_boc=90^0-1/2/_A GIVEN : A A B C in which sides A B and A C are produced to P and Q respectively. The bisectors of /_P B C and /_Q C B intersect at Odot

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 .

A B C is a triangle. The bisector of the exterior angle at B and the bisector of /_C intersect each other at D . Prove that /_D=1/2/_A .

The bisects of exterior angles at B a n d C of /_A B C meet at Odot If /_A=x^0, then /_B O C= 90^0+(x^0)/2 (b) 90^0-(x^0)/2 180^0+(x^0)/2 (d) 180^0-(x^0)/2

If O is the circumcentre of /_\ A B C and O D_|_B C , prove that /_B O D=/_A

In Figure, O is the centre of the circle and B C D is tangent to it at Cdot Prove that /_B A C+/_A C D=90^0 .

A B C is a triangle in which /_A=72^0 , the internal bisectors of angles B and C meet in O . Find the magnitude of /_B O Cdot