Question

in a parallelogram abcd , the bisectors of the consecutive angles angle A AND angle b intersect at p , show that apb is 90

Answer 1

Given :-

➦ A parallelogram ABCD

➦ AP AND BP are bisectors of consecutive angles ∠A and ∠B which intersect at P

RTP :-

➦ ∠APB = 90°

Solution :-

➦ We know, sum of the consecutive angles of a parallelogram will be supplementary (180°)

➦ ∠A + ∠B = 180°

➦ ½∠A + ½∠B = ¹⁸⁰⁄₂

➦ ∠PAB + ∠APB = 90°

➦ In ΔAPB :-

➦ ∠PAB + ∠APB + ∠PBA = 180°     (Angle sum property)

➦ ∠APB = 180° - (∠APB + ∠PBA)

➦ 180° - 90°

➦ 90°

Hence, we proved ∠APB = 90°

Answer 2

$\rm \huge \color{skyblue} ❖QuesTion:-$

In a parallelogram ABCD, the bisectors of the consecutive angles <A and <B intersect at P, show that <APB = 90°.

$\rm \huge \color{skyblue} ❖SoluTion:-$

✨Given :-

We are given a parallelogram ABCD where the bisectors of consecutive angles <A and <B intersect at P.

✨ To prove :-

We have to prove that <APB = 90°.

✨ Proof :-

⇛<A + <B = 180° [Adjacent angles are supplementary]

⇛<1 + <2 + <3 + <4 = 180°

⇛2 ( <1 + <3 ) = 180° [Angles are bisected]

⇛( <1 + <3 ) = $\dfrac{180°}{2}$

⇛<1 + <3 = 90°

✢Now, in ∆APB,

⇛<1 + <2 + <APB = 180° [Angle sum property]

⇛90° <APB = 180°

⇛<APB = 180° - 90°

⇛<APB = 90°

Hence, proved !

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