Question

state and prove of converse of midpoint theorem

Answer 1

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

Take a triangle ABC,E and F are the mid-points of side AB and AC resp.

Construction:-Through C,draw a line II BA to meet EF produced at D.

Proof:-
In Triangle AEF and CDF
1.AF=CF(F is midpoint of AC)
2.<AFE=<CFD (Vertically opp. angles)
3.<EAF=<DCF [Alt. angles,BA II CD(by construction) and AC is a transversal]
4.So,Triangle AEF = CDF(ASA)
5.EF=FD AND AE = CD (c.p.c.t)
6.AE=BE(E is midpoint of AB)
7.BE=CD(from 5 and 6)
8.EBCD is a IIgm [BA II CD (by construction) and BE = CD(from 7)]
9.EF II BC AND ED=BC (Since EBCD is a IIgm)
10.EF = 1/2 ED (Since EF = FD,from 5)
11.EF = 1/2 BC (Since ED = BC,from 9)
Hence,EF II BC AND EF = 1/2 BC which proves the mid-point theorem.


hope it helps you...

Answer 2

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)]