Question

Please solve this theoram friends..............this is class 9tg question

Answer 1

construction: draw PR//BC

∠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 = 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)

⇒ 1/2 PR = 1/2 BC

⇒ PQ = 1/2 BC [from (4)]

Answer 2

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