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The sides A B and A C of A B C are prod...

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`

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`angle7+angle3=angle2+angle6" "....(1)` (exterior antgle theorem)
Also `angle4_angle8=angle2+angle5" "....(2)` (exterior angle theorem)
Adding (1) and , we get
`angle7+angle3+angle4 angle8=angle2+angle6+angle5+angle2`
`rArr angle3+angle3+angle4 angle4=180^(@)+angle2` (`:. BO and CO angle` bisectors and by using angle sum property)

`rArr 2 angle3+2angle4=180^(@)+angle2`
Dividing both sides by
`:. angle3+angle4=90^(@)+(1)/(2)`
`rArr 180^(@)-angle1=90^(@)+(1)/(2) angle2` (by angle sum property `angle1+angle3+angle4=180^(@))`
`rArr angle=180^(@)-90^(@)-(1)/(2)angle2`
`rArr angle1=90^(@)-(1)/(2)angle2`
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