Question

prove that the straight line joining the the mid points of two sides of a triangle is parallel to the third sides​

Answer 1

Answer:

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Step-by-step explanation:

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 = 1/ 2 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 = 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)]

Thanking you.

Answer 2

Answer:

Given :- In ∆ABC, P is midpoint on AB and Q is midpoint on AC.

To prove :- PQ || BC.

Proof :- In∆ ABC, P and Q are m.p of side AB and AC, respectively.

AP = PB and AQ = QC

AP/PB = AQ/QC = 1

:-Using converse theorem of Proportionality we get,

PQ || BC. ...........(proved)

see the attachment for the required figure...