Question

State and Prove the mid point theorem

Answer 1

Heya,

STATEMENT:-

Line segments joining the mid points of sides of a triangle is parallel to the third side and half of third side.

GIVEN:-
ΔABC in which E & F are the mid points of AB and AC

TO PROVE:-
EF||BC and EF=1/2BC

CONSTRUCTION:-
Extend EF to the point D such that CD||BA

PROOF:-
ΔAEF and ΔFCD

∠A= ∠C [alternate interior angle]=>A
AF=FC [F is the mid point]=>S
∠F= ∠F [Vertically opposite angles]=>A

Therefore,
ΔAEF ≅ ΔFCD[ASA ≅]
∴ EF=FD, AE=CD [cpct]
∴BE=AE=CD
=> BE=CD, BE||CD
∴BECD is a parallelogram

=> ED=BC, ED||BC
=> EF+FD=BC, EF||BC
=> EF+EF=BC, EF||BC
=> 2EF=BC, EF||BC
=> EF= 1/2BC, EF||BC

Hope my answer helps you :)

Regards,
Shobana

Answer 2

Answer:            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

Explanation:    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)]