Question

In a ∆ABC, the sides ab and ac are produced to point d and e respectively. the bisectors of ∠DBC and ∠ECB interect at a point O .prove that ∠BOC=(90°-½∠A)​

Answer 1

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In a ∆ABC, the sides ab and ac are produced to point d and e respectively. the bisectors of ∠DBC and ∠ECB interect at a point O .prove that ∠BOC=(90°-½∠A)

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Since ABD is a line,we have

$\angle \: B + \angle \: CBD \: = 180\degree \: \: \: \: \: (linear \: pair)$

$⇒ \frac{1}{2} \angle \: B + \frac{1}{2} \angle \: CBD = 90\degree$

$⇒ \frac{1}{2} \angle \: B + \angle \: CBO = 90\degree$

$⇒ \angle \: CBO = (90\degree - \frac{1}{2} \angle \: B).$

Again,ACE is a straight line.

$\therefore \: \angle \: C + \angle \: BCE = 180\degree \: \: \: (linear \: pair)$

$⇒ \frac{1}{2} \angle \: C + \frac{1}{2} \angle \: BCE = 90\degree$

$⇒ \frac{1}{2} \angle \: C + \angle \: BCO = 90\degree$

$⇒ \angle \: BCO = (90\degree - \frac{1}{2} \angle \: C).$

We know that the sum of the angles of a triangle is 180°.

So,from ∆OBC , we get

$\angle \: CBO + \angle \: BCO \: \angle \: BOC = 180\degree$

$⇒ (90\degree - \frac{1}{2} \angle \:B) + (90\degree - \frac{1}{2} \angle \: C) + \angle \: boc = 180\degree \\ \: \: \: \: \: \: [using \: (i) \: and \: (ii)]$

$⇒ 180\degree - \frac{1}{2} (\angle \: B + \angle \: C) + \angle \: BOC = 180\degree$

$⇒ \angle \: BOC = \frac{1}{2} (\angle \: A + \angle \: B + \angle \: C) - \frac{1}{2} \angle \: A \\ \: \: \: \: \: \: [adding \: and \: subtracting \: \frac{1}{2} \angle \: A]$

$⇒ \angle \: BOC = ( \frac{1}{2} \times 180\degree) - \frac{1}{2} \angle \: A \: \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: [\therefore \: \angle \: A + \angle \: B + \angle \: C = 180\degree]$

$⇒ \angle \: BOC = (90\degree - \frac{1}{2} \angle \: A).$

$hence \: \angle \: BOc = (90\degree - \frac{1}{2} \angle \: A).$

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