In the following figure, ABC is an equilateral triangle in which QP is parallel to AC. Side AC is produced upto point R so that CR = BP.

Prove that QR bisects PC.

Given,

∆ ABC is an equilateral triangle.

∴ ∠ABC = ∠BCA = ∠CAB = 60° (Each interior angle of equilateral triangle equals to 60°.)

From figure,

⇒ ∠BPQ = ∠BCA = 60° (Corresponding angles are equal)

⇒ ∠BQP = ∠BAC = 60° (Corresponding angles)

In ∆BPQ,

By angle sum property of triangle,

⇒ ∠BQP + ∠BPQ + ∠QBP = 180°

⇒ 60° + 60° + ∠QBP = 180°

⇒ 120° + ∠QBP = 180°

⇒ ∠QBP = 180° - 120° = 60°.

Since, all the interior angles = 60°.

∴ △ BPQ is an equilateral triangle i.e., BP = PQ = BQ.

Given,

BP = CR

Since, BP = PQ.

∴ PQ = CR.

In △ MPQ and △ MCR,

⇒ ∠PQM = ∠MRC (Alternate interior angles are equal)

⇒ ∠PMQ = ∠CMR (Vertically opposite angles are equal)

⇒ PQ = CR (Proved above)

∴ ∆ MPQ ≅ ∆ MCR (By A.A.S. axiom)

We know that,

Corresponding parts of congruent triangles are equal.

∴ MP = MC

Thus, QR bisects PC.

Hence, proved that QR bisects PC.