Question

Who know the answer please do it. Please draw the picture also.​

Answer 1

Answer:

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Answer 2

Answer:Hi! I used the symbol (') to indicate degree like 90' means the angle of 90

Given : l∥m

Transversal p intersects l & m at A & C respectively. Bisector of ∠ PAC & ∠ QCA meet at B. And, bisector of ∠ SAC & ∠ RCA meet at D.

To prove : ABCD is a rectangle.

Proof :

We know that a rectangle is a parallelogram with one angle 90'

For l ∥ m and transversal p

∠PAC=∠ACR

So, 1/2<PAC=1/2<ACR

So, ∠BAC=∠ACD

For lines AB and DC with AC as transversal ∠BAC & ∠ACD are alternate angles, and they are equal.

So, AB∥DC.

Similarly, for lines BC & AD, with AC as transversal ∠BAC & ∠ACD are alternate angles, and they are equal.

So, BC∥AD.

Now, In ABCD,

AB∥DC & BC∥AD

As both pair of opposite sides are parallel, ABCD is a parallelogram.

Also, for line l,

<PAC +<CAS =180 '

1/2 <PAC +<CAS=90'

∠BAC+∠CAD=90'

∠BAD=90'

So, ABCD is a parallelogram in which one angle is 90'

.

Hence, ABCD is a rectangle.

Hope you understand