Question

Given: AX bisects angle BAC and PQ is perpendicular bisector of AC which meets AX at point Y. 

 
Prove:

1. X is equidistant from AB and AC.
2. Y is equidistant from A and C. 

Answer 1

 
Construction: From X, draw XL ⊥ AC and XM ⊥ AB. Also join YC.

Proof:

i. In ΔAXL and ΔAXM,

∠XAL = ∠XAM  ...(Given)

AX = AX  ...(Common)

∠XLA = ∠XMA  ...(Each = 90°)

∴ By Angle side angle criterion of congruence,

ΔAXL ≅ ΔAXM   ...(ASA Postulate)

The corresponding parts of the congruent triangles are congruent

∴ XL = XM  ...(C.P.C.T.)

Hence, X is equidistant from AB and AC

ii. In ΔYTA and ΔYTC,

AT = CT   ...(∵ PQ is a perpendicular bisector of AC)

∠YTA = ∠YTC   ...(Each = 90°)

YT = YT   ...(Common)

∴ By side – Angle – side criterion of congruence,

∴ ΔYTA ≅ ΔYTC   ...(SAS postulate)

The corresponding parts of the congruent triangle are congruent.

∴ YA = YC  ...(C.P.C.T.)

Hence, Y is equidistant from A and C.