Prove that opposite side mid points joining line segment is parallel to one of the parallel side.?

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Given: In triangle ABC, P and Q are mid-points of AB and AC respectively.To Prove: i) PQ || BC ii) PQ = 1 2 BCConstruction: 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 =  1 2 PR  ---------- (4)⇒ AP = CR (by CPCT)  ........(5)But, AP = BP (∵ P is the mid-point of the side AB)⇒ BP = CRAlso. BP || CR (by construction)In quadrilateral BCRP, BP = CR and BP || CRTherefore, quadrilateral BCRP is a parallelogram.BC || PR or, BC || PQAlso, PR = BC (∵ BCRP is a parallelogram)⇒ 1 2 PR = 1 2 BC⇒ PQ = 1 2 BC [from (4)]

Answer

Sorry my question is incomplete, the question is related to trapezium. But anyway thanks for the answer

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