Question

State and Prove the mid point theorem

Answer 1

Let ABC be a triangle.
Let a line segment PQ be cut through it parallel.

So, from it you can observe that
i) PQ is parallel to BC(||)
ii)PQ =1/2BC.

Therefore,
Mid-Point Theorem:-
The line segment joining the mid points of two sides of a ∆ is parallel to the third side and half of the third side.

Answer 2

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

∠1= ∠2 [alternate interior angle]=>A
AF=FC [F is the mid point]=>S
∠3= ∠4 [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