Question

Prove Mid-point theorem.

Answer 1

Given in the figure A :

AP=PB, AQ=QC.


To prove:

PQ || BC and PQ=1/2 BC


Plan:

To prove ▲ APQ ≅ ▲ QRC


Proof steps:

AQ=QC [midpoint]∠ APQ = ∠QRC [Corresponding angles for parallel lines cut by an transversal].∠PBR=∠QRC=∠APQ [Corresponding angles for parallel lines cut by an transversal].∠RQC=∠PAQ [When 2 pairs of corresponding angles are congruent in a triangle, the third pair is also congruent.]Therefore , ▲APQ ≅ ▲QRCAP=QR=PB and PQ=BR=RC.

Since midpoints are unique, and the lines connecting points are unique, the proposition is proven.

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)]