Question

In figure, ∆ABC and ∆DBC are two isosceles triangles. Prove that line segment AD lie on perpendicular bisector of base BC.

Answer 1

Given : Two isosceles triangles ABC and DBC having same base BC.

AB = AC and DB = DC

To Prove : AD lies on perpendicular bisector of base BC.

Construction : Produce AD up to point M on BC.

Proof : 

∵ ∆DBC is isosceles triangle.

∴ DB = DC

Now ∠DBC = ∠DCB

In ∆DBM and ∆DCM,

DB = DC (Given)

∠DBM = ∠DCM, (In an isosceles triangle angles  opposite to equal sides are equal).

DM = DM (Common)

∵ ∆DBM ≅ ∆DCM

∴ ∠BMD = ∠CMD and BM = CM

Now ∠BMD + ∠CMD = 180°

∠BMD = ∠CMD = 90°

∴ DM ⊥ BC

∴ M is mid-point of BC.

∴ DM is perpendicular bisector of BC.

Now point D lies on AM

Similarly, 

∴ ∆ABC is an isosceles triangle.

AM is perpendicular bisector of BC.

Thus, line segment AD, lie on ⊥ bisector of base BC.