Question

Draw any two line 'l' and 'm' and intersect in two distinct points by a transversal 'p' and write
exterior angles​

Answer 1

Answer:

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 + 1/2∠CAS = 90°

∠BAC+∠CAD=90°

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

Hence, ABCD is a rectangle.

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