Question

STATE AND PROVE MID POINT THEOREM ​PLEASE HELP ME FRIENDS

Answer 1

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)]

Answer 2

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

Step-by-step explanation:

Given:- a ∆ABC in which D and E are the mid points of sides AB and AC respectively. DE is joined.

To prove:- DE || BC and DE = 1/2 of BC.

Proof:- Produce line segment DE to F, such DE=EF. Join FC. In ∆s AED and CEF, we have

AE=CE (E is the mid point of AC)

<AED = <CEF (vertically opposite angles)

DE=EF (by construction)

So by SAS criteria for congruence ∆AED is congruent to ∆ CEF.

AD = CF ( by c.p.c.t.)

<ADE=<CFE( alt.in.angles).........(I)

Now D is the mid point of AB

AD=DB

DB=CF( from AD=CF) Now AD || FC [ from ( i )] and therefore DBCF is a parallelogram. DF || BC and DF=DF.( Opposite sides of a ||gm are equal and parallel) But, DE=EF and DE || BC.

DE =1/2 DF

therefore DE=1/2 BC ( DE = BC ).

HENCE PROVED...