State mid point theorem for class …

Mid-Point Theorem
The line segment joining the mid-points of two sides of a triangle is parallel to the third side and equal to half the third side.

Given: In triangle ABC, P and Q are mid-points of AB and AC respectively.

To Prove: i) PQ || BC ii) PQ = 1212 BC

Construction: Draw CR || BA to meet PQ produced at R.

Proof:
∠QAP = ∠QCR (Pair of alternate angles) ---------- (1)

AQ = QC (∵ Q is the mid-point of side AC) ---------- (2)

∠AQP = ∠CQR (Vertically opposite angles) ---------- (3)

Thus, ΔAPQ ≅ ΔCRQ (ASA Congruence rule)

PQ = QR (by CPCT) or PQ = 1212 PR  ---------- (4)

⇒ AP = CR (by CPCT)  ........(5)

But, AP = BP (∵ P is the mid-point of the side AB)

⇒ BP = CR

Also. BP || CR (by construction)

In quadrilateral BCRP, BP = CR and BP || CR

Therefore, quadrilateral BCRP is a parallelogram.

BC || PR or, BC || PQ

Also, PR = BC (∵ BCRP is a parallelogram)

⇒ 1212 PR = 1212 BC

⇒ PQ = 1212 BC [from (4)]