Question

∆ABC and ∆DBC are two triangles on the same base BC such that A and D lie on the opposite side of BC. AB = AC and BD = DC. Show that AD is the ⊥ bisector of BC.

Answer 1

In ∆BAD and ∆CAD

AB = AC (given)

BD = DC (given)

and AD = AD (common)

∆BAD = ∆CAD (by SSS congruence rule)

∠1 = ∠2 (by c.p.c.t)

Also AB = AC (given)

∠3 = ∠4 (angle opposite to equal sides are equal)

∆BAO = ∆CAO

⇒ BO = OC (by c.p.c.t)

or AO bisects BC (by c.p.c.t)

Also ∠AOB = ∠AOC (by c.p.c.t)

But ∠AOB + ∠AOC = 180°

⇒ ∠AOB = ∠AOC = 90°

⇒ AD is perpendicular bisector of BC.