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Question

Show that if there is a transversal between two parallel lines, the interior angle bisectors meet to form a rectangle. [4 MARKS]


Solution

Diagram : 1 Mark
Concept : 1 Mark
Proof : 2 Marks

Consider 2 parallel lines l and m.

It is given that PSQR  and a transversal  intersect them at points A and C respectively.

The bisector of PAC and ACQ intersect at B and bisectors of ACR and SAC intersect at D.

We are to show that quadrilateral ABCD is a rectangle.

Now, PAC=ACR     [Pair of alternate angles]

12PAC=12ACR 

i.e. BAC=ACD

These form a pair of alternate angles for lines AB and DC with AC as transversal and they are equal also.

AB||DC .......(i)

Similarly, BC|| AD.......(ii)    [Considering ACB and CAD]

 quadrilateral ABCD is a parallelogram.

Also, PAC+CAS=180     [linear pair]

BAC+CAD=90

BAD=90

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

ABCD is a rectangle.

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