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:

(i) F is equidistant from A and B.

(ii) F is equidistant from AB and AC.

Answer 1

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.

Construction: Draw LF ⊥ AC

Proof: In Δ AFL and ΔAFE,

∠FEA = ∠FLA (Each = 90°)

∠LAF = FAE (AD ** 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.