Math, asked by alaricboss, 11 months ago

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​

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Answered by Anonymous
132

\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)


BrainlyPromoter: Fantastic answer :)
Answered by Blaezii
42

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°
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