Question

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Answer 1

✰✰ ANSWER ✰✰

➮Given : -

• BO Is bisector of $\angle{CBE}$
• CO is bisector of $\angle{BCD}$

➮To Prove : -

• $\implies\tt{\angle{BOC}=90\degree-\dfrac{1}{2}\angle{BAC}}$

➮Proof : -

$\implies\tt{\angle{ABC}+\angle{CBE}=180\degree}$ (Linear Pair)

$\implies\tt{y+\angle{CBE}=180\degree}$

$\implies\tt{\angle{CBE}=180-y}$

→BO is Bisector of $\angle{CBE}$ : -

$\implies\tt{\angle{CBO}=\dfrac{180-y}{2}\:and\:\angle{OBE}=\dfrac{180-y}{2}}$

➮Similarly : -

$\implies\tt{\angle{BCO}=\dfrac{180-z}{2}\:and\:\angle{DCO}=\dfrac{180-z}{2}}$

➥Now , In Δ BOC : -

$\implies\tt{\angle{CBO}+\angle{BCO}+\angle{BOC}=180\degree}$ (A.S.P)

$\implies\tt{\dfrac{180-y}{2}+\dfrac{18p-z}{2}+\angle{BOC}=180\degree}$

$\implies\tt{\dfrac{180}{2}-\dfrac{y}{2}+\dfrac{180}{2}-\dfrac{z}{2}+\angle{BOC}=180\degree}$

$\implies\tt{90-\dfrac{y}{2}+\dfrac{90-z}{2}+\angle{BOC}=180\degree}$

$\implies\tt{180-\dfrac{y}{2}-\dfrac{z}{2}+\angle{BOC}=180\degree}$

$\implies\tt{\angle{BOC}=180-180+\dfrac{y}{2}+\dfrac{z}{2}}$

$\implies\tt{\angle{BOC}=\dfrac{y+z}{2}}$ -----(1)

➮Now , In Δ ABC : -

$\implies\tt{x+y+z=180\degree}$ (Angle Sum Property)

$\implies\tt{y+z=180-x}$ -----(2)

➥By Equation 1 and 2 : -

$\implies\tt{\angle{BOC}=\dfrac{y+z}{2}-\dfrac{180-x}{2}}$

$\implies\tt{\angle{BOC}=\cancel\dfrac{180}{2}-\dfrac{x}{2}}$

$\implies\tt{\angle{BOC}=90-\dfrac{x}{2}}$

$\implies\tt{\angle{BOC}=90-\dfrac{1}{2}\angle{BAC}}$

✺ HENCE PROVED ✺