Question

In fig Side AB and AC of triangle ABC are produced to E and D respectively .If angle bisector BO and CO of angle CBE and angle BCD meet aech other at Point O , then prove that : angle BOC = 90 degree - angle x /2

Answer 1

$\huge{\bf{\underline{\underline{\purple{Solution :}}}}}$

Given,

• Bisectors BO and CO of ∠CBE and ∠BCD meet each other at point O.

To Proof :

• $\bold{\angle BOC = 90^{\circ} - \frac{\angle x}{2}}$

Proof :

$\implies \bold{\angle EBC + y = 180^{\circ}}$

$\implies \bold{\angle EBC = 180^{\circ} - \angle y}$

$\bold{\therefore \angle OBC = \frac{1}{2} \angle EBC}$

⇒ ∠OBC = 1/2  (180° - ∠y)..............2) Equation

$\implies \bold{\angle OCB + \angle z = 180^{\circ}}$

$\implies \bold{\angle DCB = 180^{\circ} - \angle z }$

$\implies \bold{\therefore \angle OCB = \frac{1}{2} \angle DCB }$

⇒ ∠OCB = 1/2 ( 180° - ∠z)...........................2) Equation

$\rule{200}2$

$\implies \bold{\angle BOC + \angle OBC + \angle OCB = 180^{\circ}}$

$\implies \bold{\angle BOC + \frac{1}{2} (180^{\circ} - \angle y ) +\frac{1}{2} (180^{\circ} - \angle z) = 180^{\circ}}$

$\implies \bold{\angle BOC + \frac{1}{2} ( 180^{\circ} - \angle y + 180^{\circ} - \angle z ) = 180^{\circ}}$

$\implies \bold{2\angle BOC + 360^{\circ} - \angle y - \angle z = 360^{\circ}}$

$\implies \bold{2\angle BOC = \angle y + \angle z }$

[ Sum of all angle of triangle = 180°

⇒∠x + ∠y +∠z = 180°

⇒∠y + ∠z = 180° - x ]

Now put value of ∠y + ∠z :

$\implies \bold{\angle BOC = 180^{\circ} - \angle x}$

$\implies \bold{\angle BOC = \frac{180^{\circ} - \angle x }{2} }$

$\implies \boxed{\bold{\angle BOC = 90^{\circ} - \frac{\angle x}{2}}}$

Hence Proved

$\rule{200}2$

Answer 2

$\huge\mathbb {SOLUTION:-}$

$\setlength{\unitlength}{1.0 cm}}\begin{picture}(12,4)\thicklines\put(1,1){\line(1,0){6.5}}\put(1,1.1){\line(1,0){6.5}}\end{picture}$

∠CBE = 180 - ∠ABC

∠CBO = $\mathsf {\frac{1}{2} }$ ∠CBE

• (BO is the bisector of ∠CBE)

∠CBO = $\mathsf {\frac{1}{2} }$ ( 180 - ∠ABC) $\mathsf {\frac{1}{2} }$ x 180 = 90

∠CBO = 90 - $\mathsf {\frac{1}{2} }$∠ABC .............(1)

$:\implies \mathsf {\frac{1}{2} }$x ∠ABC = $\mathsf {\frac{1}{2} }$∠ABC

$:\implies$∠BCD = 180 - ∠ACD

$:\implies$∠BCO = $\mathsf {\frac{1}{2} }$∠BCD

• ( CO is the bisector os ∠BCD)

$:\implies$∠BCO = $\mathsf {\frac{1}{2} }$ (180 - ∠ACD)

$\setlength{\unitlength}{1.0 cm}}\begin{picture}(12,4)\thicklines\put(1,1){\line(1,0){6.5}}\put(1,1.1){\line(1,0){6.5}}\end{picture}$

∠BCO = 90 - $\mathsf {\frac{1}{2} }$∠ACD .............(2)

$:\implies$∠BOC = 180 - (∠CBO + ∠BCO)

$:\implies$∠BOC = 180 - (90 - $\mathsf {\frac{1}{2} }$∠ABC + 90 - $\mathsf {\frac{1}{2} }$∠ACD)

$:\implies$∠BOC = 180 - 180 + $\mathsf {\frac{1}{2} }$∠ABC + $\mathsf {\frac{1}{2} }$∠ACD

$:\implies$∠BOC = $\mathsf {\frac{1}{2} }$ (∠ABC + ∠ACD)

$:\implies$∠BOC = $\mathsf {\frac{1}{2} }$ ( 180 - ∠BAC) (180 - ∠BAC = ∠ABC + ∠ACD)

$:\implies$∠BOC = 90 - $\mathsf {\frac{1}{2} }$∠BAC

Hence proved.........

$\setlength{\unitlength}{1.0 cm}}\begin{picture}(12,4)\thicklines\put(1,1){\line(1,0){6.5}}\put(1,1.1){\line(1,0){6.5}}\end{picture}$