Question

Two parallel lines l and m are intersected by a transversal p. Show that the quadrilateral formed by the bisectors of interior angles is a rectangle.​

Answer 1

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 90o.

For l∥m and transversal p

∠PAC=∠ACR

So, 21∠PAC=21∠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=180o

21∠PAC+21∠CAS=90o

∠BAC+∠CAD=90o

∠BAD=90o.

So, ABCD is a parallelogram in which one angle is 90o.

Hence, ABCD is a rectangle.

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Answer 2

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Step-by-step explanation:

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