In the given figure on the common base BC there are two isosceles triangles ΔPBC and ΔQBC on the opposite sides of BC. Prove that the line joining the points P and Q bisects line BC at right angles.

In the given figure ∆PBC and ∆QBC are isosceles triangles.

So, PB=PC and BQ=QC

To prove: line joining the points P and Q bisects line BC at right angles.

Proof: Now, in ∆PBQ and ∆PCQ

PB=PC (given)

BQ=QC (given)

PQ=PQ (common)

∴ ∆PBQ ∆PCQ (by SSS rule)

∠BPO=∠CPO (by cpct)………(1)

And ∠BQO=∠CQO (by cpct)…………(2)

Now in ∆BOP and ∆COP,

BP=CP (given)

OP=OP (common)

∠BPO=∠CPO (from (1))

∴ ∆BOP ∆COP (by SAS rule)

So, BO=CO (by cpct)

∠BOP=∠COP (by cpct)………(3)

Now, ∠BOP+∠COP=180°

⇒ ∠BOP+∠BOP=180° (from (3))

⇒ 2∠BOP = 180°

⇒ ∠BOP = 90°

Hence , ∠BPO=∠CPO=90°

∴ PQ bisects BC at right angles.