Question

the sides AB and AC of triangle ABC are produed to point E and D respectively .if bisectors BO and CO of angle CBE and angle BCD respectively meet at point O, then prove that angle BOC=90 - half of angle BAC​

Answer 1

$\huge\boxed{Answer:-}$

• Let BO the Ray and also BO is the bisector of CBE

Hence, CBO = Half CBE(1/2)

$= \: \frac{1}{2} (180°-y) \\ </p><p>=90°-y/2 equation \: (1)$

Let CO the Ray and also the bisector of BCD

Hence, BCO = Half BCD(1/2

$= \: \frac{1}{2} (180°-z )\\ </p><p>= \: 90° \frac{ - z}{2} \: equation \: \: (2)$

The Third equation:-

Adding all the triange we get 180°

BOC and BOC + BCO+ CBO=180° _equatiob - (3)

Adding substituting for equation 1,2 and 3

$(BOC + 90° \frac{ - z}{2} +90° \frac{ - z}{2} =180°)</p><p>$

$(BOC= \frac{z}{2} + \frac{y}{2} )$

and ,

$BOC= \frac{1}{2}...z+y.. Equation - (4)$

After adding all the angle we get 180°

$(X + Y + Z = 180° ) \\ </h3><h3>(y + z=180°-x)$

Hence,equation (4)

$(BOC = 12.....180°-x)$

$=(90° \frac{ - x}{2} )</p><p>$

$</p><p>=90° \frac{ - 1}{2} BAC$

Property used to solve this question:-

ASP (Angle Sum property)

Answer 2

Answer :

Proved.

Step-by-step explanation:

Given :

The sides AB and AC of triangle ABC are produed to point E and D.

Bisectors BO and CO of ∠CBE and ∠BCD respectively meet at point O.

To Prove :

$\bf\\\angle\;BOC = 90 - \dfrac{1}{2}\;\angle\;BAC$

$\bigstar$ Prove :

BO is the bisector of ∠CBE

So, ∠BCO  = ∠DCO  = $\dfrac{1}{2}$

∠ CBE is the exterior angle of Δ ABC.

Implies,

$\sf \angle\;CBE = x+y$

⇒ $\dfrac{1}{2}$ ( ∠CBE ) = $\dfrac{1}{2}$ (x + y)

⇒ ∠ CBO = $\dfrac{1}{2}$ (x + y)

Same here,

∠BCD is the exterior angle of Δ ABC

So,

⇒ ∠ BCD = (a + b).

⇒ $\dfrac{1}{2}$ ( ∠BCD ) = $\dfrac{1}{2}$ (a + b)

⇒ ∠ BCD = $\dfrac{1}{2}$ (a + b)

In ΔOBC,

⇒ ∠ BOC  + ∠ BCO + CBO = 180°

★__________{ Angle Sum Property }

⇒ ∠ BOC $\dfrac{1}{2}$ ( a + b) + $\dfrac{1}{2}$ (x + y) = 180°

⇒ ∠ BOC + $\dfrac{1}{2}$ ( a + b + x + y) = 180°    .....Eq(1)

In Δ ABC

( a + b + x + y) = 180°    .....Eq(2)

★__________{ Angle Sum Property. }

Putting values of Eq(2) in Eq(1),

⇒ ∠ BOC + $\dfrac{1}{2}$ (a + b + x + y) = 180°

⇒ ∠ BOC + $\dfrac{1}{2}$  ( a + 180°) = 180°

⇒ ∠ BOC + $\dfrac{a}{2}$ + $\dfrac{1}{2}$ × 180°

⇒ ∠ BOC = $\dfrac{a}{2}$ + 90°  = 180°

⇒ ∠ BOC = 180° - 90° = $\dfrac{x}{2}$

⇒ ∠ BOC = 90° - $\dfrac{1}{2}$

⇒ ∠ BOC = 90° - $\dfrac{1}{2}$ × ∠ BAC

Hence, Proved!

$\rule{300}{1.5}$

$\bigstar$ Points to Remember -

• External angle is sum of two interior opposite angles.
• Angle Sum Property, i.e - x + y + z = 180°