Question

ABC and DBC are two triangles on the same base BC such that A and D lie on the opposite sides of BC, AB = AC and DB = DC. Show that AD is the perpendicular bisector of BC.

Answer 1

Given in the question, ABC and DBC are two triangles on the same base BC such that A and D lie on the opposite sides of BC, AB = AC and DB = DC.

To proof that AD is the perpendicular bisector of BC that is OB = OC.

Proof: In triangle BAD and triangle CAD,

AB = AC  ...[Given]

BD = CD  ...[Given]

AD = AD  ...[Common side]

Now, by SSS criterion of congruence,

ΔBAD ≅ ΔCAD

So, ∠1 = ∠2   ...[CPCT]

Now, in triangle BAO and triangle CAO,

AB = AC  ...[Given]

∠1 = ∠2  ...[Proved above] 

AO = AO   ...[Common side]

So, by SAS criterion of congruence,

ΔBAO ≅ ΔCAO

Since, BO = CO  ...[CPCT]

And ∠3 = ∠4  ...[CPCT]

∠3 + ∠4 = 180°   ...[Linear pair axiom]

∠3 + ∠3 = 180°

2∠3 = 180°

∠3 = `(180^circ)/2`

∠3 = 90°

Therefore, AD is perpendicular to bisector of BC.

Hence proved.