Question

hw to solve this mam? ​

Answer 1

Answer:

See the pic you will get your answer.

Answer 2

$\tt\it\bf\it\Huge\bm{\mathcal{\fcolorbox{red}{orange}{\red{ANSWER:-}}}}$

☺️☺️

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$\tt\it\bf\it\LARGE\bm{\mathcal{\fcolorbox{red}{white}{\blue{Given:-}}}}$

➠ ABCD is $\tt\Large{\red{ parallelogram}}$

➠Angle bisector of B and C intersct at P

$\tt\it\bf\it\Large\bm{\mathcal{\fcolorbox{blue}{white}{\orange{To\:Prove:-}}}}$

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➠ ∠BPC= 90°

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$\tt\it\bf\it\LARGE\bm{\mathcal{\fcolorbox{blue}{white}{\pink{Formula\:to\:be\:Used:-}}}}$

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➠ Angle sum property:-

For, ▲ ABC

 ∠A +  ∠B +  ∠C = 180°

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$\tt\it\bf\it\Huge\bm{\mathcal{\fcolorbox{blue}{white}{\green{Solution:-}}}}$

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Since, ABCD is a $\tt</strong><strong>\</strong><strong>L</strong><strong>a</strong><strong>r</strong><strong>g</strong><strong>e</strong><strong>{\red{ parallelogram}}$

•°• AD//BC

BC is transversal.

•°•  ∠B +  ∠C = 180°

Multiplying both sides by 1/2:-

1/2 (  ∠B +  ∠C) = 1/2(180°)

=>1/2 ∠B + 1/2 ∠C = 90°------(1)

Since, BP and PC are the bisector of  ∠B and  ∠C

•°•  ∠BCP = 1/2  ∠C

•°•  ∠CBP = 1/2  ∠B

Substitute the value in eqn(1):-

We get,

 ∠BCP +  ∠CBP= 90°

Now,

In ▲BPC:-

 ∠BCP +  ∠CBP + ∠BPC = 180°

=>90° +  ∠BPC =180°

=> ∠BPC =180°-90°

=> ∠BPC =90°

$\tt\it\bf\it\LARGE\bm{\mathcal{\fcolorbox{blue}{white}{\red{Hence,\:proved}}}}$

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$\tt{\mathcal\red{ Thank\:you }}$✌️✌️