Mathematics
Given: CP is the bisector of angle C of ∆ABC.
Prove: P is equidistant from AC and BC.
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Answer
From P, draw PL ⊥ AC and PM ⊥ CB
In ∆LPC and ∆MPC,
⇒ ∠PLC = ∠PMC [Each 90°]
⇒ ∠PCL = ∠MCP [Since, CP is bisector of angle C]
⇒ PC = PC [Common]
∴ ∆LPC ≅ ∆MPC by AAS axiom.
∴ PL = PM [By C.P.C.T.]
Hence, proved that P is equidistant from AC and BC.
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