Question

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Please solve it!

Answer 1

Step-by-step explanation:

Given :

A parallelogram ABCD in which bisectors of angles A, B, C, D intersect at P, Q, S to form a quadrilateral PQRS.

To Proof :

Each angle of PQRS is a right angle i.e. it's a rectangle.

Solution :

We know, AB is a llgm.

So - AB II DC, AB II DC and AD intersects at D and A.

Hereby, we get,

∠A + ∠D = 180° (sum of interior angles)

$\displaystyle{ \sf{\frac{1}{2} \angle \: A \: + \frac{1}{2} \angle \:D = 90 {}^{ \circ} }}$

∠DAS + ∠ADS = 90° - eq.1

(DS and AS are bisectors of ∠A and ∠D)

Now, in ∆DAS :

∠DAS + ∠ASD + ∠ADS = 180°

( sum of angle in a triangle)

90° + ∠ASD = 180° (from eq-1)

∠ASD = 180° - 90°

∠ASD = 90°

∠ASD = 90° = ∠PSR

(vertically opposite angles)

Similarly, we can prove :

∠RQP = 90°

∠SRQ = 90°

∠SPQ = 90°

Therefore, each angles of PQRS is right angle and it can be called rectangle.

Note:

Picture is also attached.

Answer 2

Answer:

Given:

ABCD is a parallelogram where AB parallel to CD

and AD parallel to BC

To prove :

all the angles of quadrilateral pqrs is 90 degree or right angle

prove :

we know that if a transverse line cuts two parallel lines then the sum of interior angles on the same side of the line is 180 degree

so

angle D + angle A = 180 degree. _______(1)

here BQ is a bisector of angle D and AS is bisector of angle A

so

: angle PDA=✓D/2

and

angle PAD =✓A/2

now from equation 1

we get

(✓D+✓A)/2=180°

or

✓D/2+✓A/2=90°

Now

in triangle ADP

✓SPQ=✓PDA+✓PAD. ( exterior angle theorem)

: ✓SPQ=90°

similarly we can prove other three angles

✓PSR=90°

✓SRQ=90°

✓RQP=90°

therefore each angle of quadrilateral pqrs is a right angle.

Hence proved

hope it helps you ✌️✅✅✅