Question

How to prove mid point theorem

Answer 1

Answer:

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.

Step-by-step explanation:

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

Mid point Theorem :

The line segment joining the mid points of any two sides of a triangle is parallel to the third side.

Given :

A \triangle ABC△ABC in which D and E are the mid points of AB and AC, respectively.

To prove :

DE \parallel BCDE∥BC.

Proof :

Since D and E are the mid points of AB and AC, respectively, we have AD=DBAD=DB and AE=ECAE=EC.

Therefore,

\dfrac{AD}{DB}=\dfrac{AE}{EC}  

DB

AD

​  

=  

EC

AE

​  

           ( each equal to 1 )

Therefore, by the converse of thales theorem, DE \parallel BCDE∥BC.