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Question
In Δ PQR, s is a point on PR such that ∠ PQS = ∠ RQS . Prove thats is equidistant from PQ and QR.
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Solution
Steps of Construction:
(i) Draw line segment PQ.
(ii) With P and Q as centre draw intersecting arcs at R.
(iii) Join PR and RQ.
(iv) Draw angle bisector of angle Q.
(v) Draw perpendicular bisectors of PQ and RQ which meet the angle bisector at S. S is the required point.
(vi) In Δ QSY and Δ QSX
SQ= SQ
∠ SQY = ∠ SQX
∠ SYQ = ∠ SXQ = 90 degrees.
Therefore, Δ QSY and Δ QSX are congruent.
Hence, SY = SX and therefore S is equidistant from PQ and RQ.
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