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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Answers
Given : a straight line AB intersects two parallel lines CD and EF at two distinct points P and Q respectively.
To find : prove that internal bisectors of ∠DPQ and ∠PQF meet each other at right angles.
Solution:
Let say internal bisectors of ∠DPQ and ∠PQF meet each other at R
Need to show that ∠QRP = 90°
in Δ PQR
∠QPR + ∠PQR + ∠QRP = 180° ( sum of angles of triangle)
PR is angle bisector of ∠DPQ => ∠QPR = ∠DPQ/2
& QR is angle bisector of ∠PQF => ∠PQR = ∠PQF /2
=> ∠DPQ/2 + ∠PQF /2 + ∠QRP = 180°
=> (∠DPQ + ∠PQF)/2 + ∠QRP = 180°
∠DPQ + ∠PQF = 180° as CD ║ EF & AB intersect it at P & Q
=> 180° /2 + + ∠QRP = 180°
=> 90° + ∠QRP = 180°
=> ∠QRP = 90°
QED
Hence proved
∠DPQ and ∠PQF meet each other at right angle
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