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Question
In the given figure ABC is a triangle. CP bisects angle ACB and MN is perpendicular bisector of BC. MN cuts CP at Q. Prove Q is equidistant from B and C, and also that Q is equidistant from BC and AC.
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Solution
Join BQ and draw perpendicular bisector of AC cutting AC at L.
In Δ QBN and ΔQCN
QN = QN
BN =NC
∠ QNB = ∠ QNC = 90 degree.
Therefore, ∠ QBN and ∠.QCN are congruent .
Hence Q is equidistant from B and C.
In Δ QNC and Δ QLC
QC= QC
∠ QLC = ∠ QNC = 90 degree.
∠ QCL =∠ QCN (PC being angle bisector)
Therefore, .Δ QNC and Δ QLC are congruent.
Therefore, QL = QN.
Hence Q is equidistant from BC and AC.
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