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Question
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.
Prove that:
F is equidistant from AB and AC.
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Solution
Construction: Draw LF ⊥ AC
Proof: In ΔAFL and ΔAFE,
∠FEA = ∠FLA ...(Each = 90°)
∠LAF = FAE ...(AD bisects ∠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 ...(C.P.C.T.)
Hence, F is equidistant from AB and AC.
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