Question

Suppose a straight line AB intersects two parallel lines CD and EF at two distinct points P and Q respectively. Prove that the internal bisectors of angle DPQ and angle PQF meet each other at right angles.
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Answer 1

Given: Suppose a straight line AB intersects two parallel lines CD and EF at two distinct points P and Q respectively.

To find: Prove that the internal bisectors of angle DPQ and angle PQF meet each other at right angles.

Solution:

• Now we have given that internal bisectors meet each other, so let the point be R.

• Consider the triangle PQR, we have:

                ang QPR + ang PQR + ang QRP = 180   (ASP of triangle)

• Now QR is angle bisector of ang PQF so

                ang PQR  = ang PQF /2

• Also PR is angle bisector of angDPQ so:

                ang QPR = ang DPQ/2

• Now:

                ang DPQ/2 + ang PQF /2 + ang QRP = 180

                (ang DPQ + ang PQF)/2 + ang QRP = 180

                ang DPQ +  ang PQF = 180  

• Because CD||EF and AB intersect it at P & Q.

                180 /2 +  + ang QRP = 180

                ang QRP = 90

• Hence proved

Answer:

            So we proved that angle DPQ and angle PQF  meet each other at right angle