Question

please answer it correctly and fast. very important​

Answer 1

Here BO,CO are the angle bisectors of ∠EBC & ∠DCB intersect each other at O.

∴∠EBO=∠OBC and ∠OCB=∠OCD

Side AB and AC of △ABC are produced to E and D respectively.

∴∠EBC=∠BAC+∠ACB --------------(1)

And ∠DCB=∠BAC+∠ABC --------------(2)

Adding (1) and (2) we get

∠EBC+∠DCB=2∠BAC+∠ABC+∠ACB

2∠OBC+2∠OCB=∠BAC+180 ∘

------Sum of interior angles of a triangle is equal to 180 ∘

∠OBC+∠OCB= 21

∠A+90 ∘

----------(3)

But in a △BOC=∠OBC+∠OCB+∠BOC=180∘--------(4)

From equations (3) and (4) we get 21

∠BAC+90 ∘

+∠BOC=180 ∘

∠BOC=90 ∘ − 21

∠BAC

Answered By=shreyash. ..

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Answer 2

$\longmapsto\:\:\bf{\angle{CBE}\:=\:180°\:-\:\angle{y}\:}$

$:\implies\:\:\bf{\angle{CBO}\:=\:\dfrac{1}{2}\:\angle{CBE}\:}$

$:\implies\:\:\bf{\angle{CBO}\:=\:\dfrac{1}{2}\:(180°\:-\:y)\:}$

$:\implies\:\:\bf{\angle{CBO}\:=\:90°\:-\:\dfrac{y}{2}\:}----(i)$

$\Large\bf\red{Again,}$

$\longmapsto\:\:\bf{\angle{BCD}\:=\:180°\:-\:\angle{z}\:}$

$:\implies\:\:\bf{\angle{BCO}\:=\:\dfrac{1}{2}\:\angle{BCD}\:}$

$:\implies\:\:\bf{\angle{BCO}\:=\:\dfrac{1}{2}\:(180°\:-\:z)\:}$

$:\implies\:\:\bf{\angle{BCO}\:=\:90°\:-\:\dfrac{z}{2}\:}----(ii)$

$\longrightarrow\:\:\bf{\angle{BOC}\:=\:180°\:-\:\Big(\angle{CBO}\:+\:\angle{BCO}\Big)\:}$

$\longrightarrow\:\:\bf{\angle{BOC}\:=\:180°\:-\:\Big(90°\:-\:\dfrac{1}{2}{y}\:+\:90°\:-\:\dfrac{1}{2}z\Big)\:}$

$\longrightarrow\:\:\bf{\angle{BOC}\:=\:\dfrac{y}{2}\:+\:\dfrac{z}{2}\:}$

$\longrightarrow\:\:\bf{\angle{BOC}\:=\:\dfrac{1}{2}\:(y\:+\:z)\:}$

$\longrightarrow\:\:\bf{\angle{BOC}\:=\:\dfrac{1}{2}\:(180°\:-\:x)\:}$

$\longrightarrow\:\:\bf{\angle{BOC}\:=\:90°\:-\:\dfrac{1}{2}x\:}$

$\longrightarrow\:\:\bf\green{\angle{BOC}\:=\:90°\:-\:\dfrac{1}{2}{\angle{BAC}}\:}\:[Hence\: Proved]$