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.
F is equidistant from A and B.
Construction: Join FB and FC
Proof: In ΔAFE and ΔFBE,
AE = EB (E is the mid-point of AB)
∠FEA = ∠FEB (Each = 90°)
FE = FE (Common)
∴ By side Angle side criterion of congruence,
ΔAFE ≅ ΔFBE (SAS Postulate)
The corresponding parts of the congruent triangles are congruent.
∴ AF = FB (CPCT)
Hence, F is equidistant from A and B.
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