Question

ABCD is a parallelogram.AP,BP,RD and RC are the angle bisectors of angles a,b,c and d respectively.Prove that PQRS is a parallelogram.
Pls answer it quickly it’s urgent

Answer 1

Correct Question -

ABCD is a parallelogram.AP,BP,RD and RC are the angle bisectors of angles A, B, C and D respectively.Prove that PQRS is a rectangle.

(As, bisectors of parallelogram forms rectangle)

Solution -

Given:

• ABCD is a parallelogram.
• AP, BP, RD and RC are the bisectors of $\sf{\angle{A}}$,$\sf{\angle{B}}$, $\sf{\angle{C}}$ and $\sf{\angle{D}}$

To Prove:

• PQRS is a rectangle.

Proof:

[ Refer the attachment for figure ]

As, we know that opposite sides of parallelogram are parallel. Then,

$\implies\:\sf{AB\:||\:DC}$

And AD is transversal on parallel lines AB and DC. Then,

$\implies\:\sf{\angle{A}\:+\:\angle{D}\:=\:180^{\circ}}$

Half of sum of $\sf{\angle{A}}$ and $\sf{\angle{D}}$ is equal to 90°

i.e.

$\implies\:\sf{\frac{1}{2}\angle{A}\:+\:\frac{1}{2}\angle{D}\:=\:\frac{1}{2}180^{\circ}}$

$\implies\:\sf{\frac{1}{2}\angle{A}\:+\:\frac{1}{2}\angle{D}\:=\:90^{\circ}}$ ...(1)

Means, $\sf{\angle{DAS}\:+\:\angle{ADS}\:=\:90^{\circ}}$ ...(2)

In ∆ADS

⇒ $\sf{\angle{A}\:+\:\angle{D}\:+\:\angle{S}\:=\:180^{\circ}}$

⇒ $\sf{90^{\circ}\:+\:\angle{S}\:=\:180^{\circ}}$ [From (1)]

⇒ $\sf{\angle{S}\:=\:90^{\circ}}$

($\sf{\angle{DSA}}$ = 90°)

Now, AP and DR intersect each other at point S.

$\implies\:\sf{\angle{PSR}\:=\:\angle{DSA}}$

$\implies\:\sf{PSR\:=\:90^{\circ}}$

Similarly, $\sf{\angle{SRQ}}$, $\sf{\angle{RQP}}$ and $\sf{\angle{QPS}}$ = 90°

$\implies\:\sf{\angle{PSR}\:=\:\angle{PQR}}$

Also,

$\implies\:\sf{\angle{SRQ}\:=\:\angle{QPS}}$

From above calculations we have -

⇒ $\sf{\angle{PSR}}$ = $\sf{\angle{SRQ}}$ = $\sf{\angle{RQP}}$ = $\sf{\angle{QPS}}$ = 90°

Opposite angles of PQRS are equal means PQRS is a parallelogram. But one of the angle is 90° means PQRS is a rectangle.

Answer 2

$\huge\tt\blue{Answer}$

$\tt\red{Given}$

____________________________________________

1) ABCD is a parallelogram.

2) AP,BP,RD and RC are the angle bisectors of angles a,b,c and d respectively.

3) Prove that PQRS is a parallelogram.

____________________________________________

$\tt\red{→AB || DC}$

$\tt\red{→∠A + ∠D = 180°}$

$\tt\red{→∠A, ∠D = 90°}$

____________________________________________

$\tt\green{→ \frac{1}{2} \: ∠A \: + \: \frac{1}{2} \: ∠D \: = \: \frac{1}{2} \: 180°}$

$\tt\green{→∠A \: + \: \frac{1}{2} \: \: ∠D \: = \: 90° \: \: \: \: \: \: \: \: \: \: \: \: - (1)}$

$\tt\green{→∠DAS \: + \: ∠ADS \: = \: 90° \: \: \: \: \: \: \: \: \: \: \: \: - (2)}$

___________________________________________

∆ ADS

$\tt\red{→∠A + ∠D + ∠S = 180°}$

$\tt\red{→90° + ∠S = 180° \: \: \: \: \: \: \: \: \: \: \: - (1)}$

$\tt\red{→∠S = 90°}$

$\tt\red{→∠SA = 90°}$

$\tt\red{→∠PSR = ∠DSA}$

$\tt\blue{→ ∠PSR = 90°}$

____________________________________________

$\tt\blue{→∠SRQ, ∠RQP, ∠QPS = 90°}$

$\tt\blue{→∠PSR = ∠PQR }$

$\tt\blue{→∠SRQ = ∠QPS}$

____________________________________________

$\tt\green{→∠PSR = ∠SRQ = ∠RQP = ∠QPS = 90°}$

As, opposite sides are equal and one angle is 90° , so it's a rectangle.

$\huge\tt\red{Hence \:Proved}$