The given figure shows a triangle ABC in which AD bisects angle BAC. EG is perpendicular bisector of side AB which intersects AD at point F.
(ii) F is equidistant from AB and AC.
Construction: Draw LF ⊥ AC
Proof: In Δ AFL and ΔAFE,
∠FEA = ∠FLA (Each = 90°)
∠LAF = FAE (AD bi sects ∠BAC)
AF = AF (common)
∴ By angle – Angle side criterion of congruence,
ΔAFL ≅ AFE (AAS postulate)
The corresponding parts of the congruent triangles are congruent.
∴ FE = FL (CPCT)
Hence, F is equidistant from AB and AC.
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