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Question
Given: CP is bisector of angle C of ΔABC.
Prove: P is equidistant from AC and BC.
Solution
Construction: From P, draw PL ⊥ AB and PM ⊥ CB
Proof: In ΔLPC and ΔMPC,
∠PLC = ∠PMC ...(Each = 90°)
∠PCL = ∠MCP ...(Given)
PC = PC ...(Common)
∴ By angle- side angle criterion of congruence,
ΔLPC ≅ ΔMPC ...(AAS postulate)
The corresponding parts of the congruent triangles are congruent
∴ PL = PM ...(C.P.C.T.)
Hence, P is equidistant from AC and AB
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