Question

please solve following questions..​

Answer 1

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In a $\triangle\:ABC$,the sides AB and AC are produced to point D and E respectively.The bisectors of $\angle\:DBC$ and $\angle\:ECB$ intersect at a point O.

To Prove

$\sf\angle\:BOC=\bigg(90^{\circ}-\dfrac{1}{2}\angle A\bigg)$

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$\bold{Explanation:}$

Given:

OB and OC are angle bisector of DBC and ECB.

$\implies\sf\angle\:OBC = \dfrac{1}{2} \angle \: DBC$.........(1)

Similarly,

$\implies\sf\angle\:OCB= \dfrac{1}{2} \angle \: ECB$..........(2)

$\implies\sf\angle\:DBC= \angle \: A+\angle\:ACB$....…...(3)[exterior angle property]

$\implies\sf\angle\:ECB = \angle \:A+\angle\:ABC$........(4)[exterior angle property]

In $\sf\triangle\:BOC$

$\implies\sf\angle\:BOC+\angle\:OBC+\angle\:OCB= 180^{\circ}$[Angle sum property]

$\implies\sf\angle\:BOC = 180^{\circ}-\angle\:OBC-\angle\:OCB$

$\implies\sf 180^{\circ}-\bigg(\angle\:OBC+\angle\:OCB\bigg)$.........(5)

By Using Equation 1,2,3 and 4 we have ,

$\implies\sf\angle\:OBC = \dfrac{1}{2} \angle \: DBC=\dfrac{1}{2}\bigg(\angle A+\angle\:ACB\bigg)$

$\implies\sf\angle\:OCB = \dfrac{1}{2} \angle \: ECB=\dfrac{1}{2}\bigg(\angle A+\angle\:ABC\bigg)$

$\implies\sf\angle\:BOC =180^{\circ}-\bigg(\angle\:OBC+\angle\:OCB\bigg)$

$\implies\sf 180 =\bigg[\dfrac{1}{2} \bigg(\angle A+ \angle\:ACB\bigg) + \dfrac{1}{2}\bigg (\angle A+ \angle\:ABC\bigg)\bigg]$

$\implies\sf 180 - \bigg[ \dfrac{1}{2} \bigg(\angle A\bigg) + \dfrac{1}{2} \bigg(\angle A+ \angle\:ABC + \angle\:ACB\bigg)\bigg]$

$\implies\sf 180 -\bigg ( \dfrac{1}{2} \angle A + \dfrac{180^{\circ}}{2}\bigg )$

$\implies\sf\angle\:BOC= 90 - \dfrac{1}{2} \angle A\bigg(Proved\bigg)$