Question

Given: CP is bisector of angle C of ΔABC. 

Prove: P is equidistant from AC and BC. 

Answer 1

 
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