Question

Please do it in copy.....

Answer 1

Given : ABCD is a parallelogram . L is the mid point of AB ; AL = LB

To Prove : ACBM is a parallelogram

Construction : Join AC and MB which form a quadrilateral ACBM

Proof:

AD || CB ........( ABCD is a parallelogram)

AD+AM || CB

AM || CB .....( A)

In ∆ ALM & ∆ CLB

$\angle{4} = \angle{3}$ ....(alt. int. angle)

$AL \: = \: LB$ ......(L is mid point of AB)

$\angle{1} = \angle{2}$ ...(V.O.A property)

Triangles are congruent by ASA property.

ML = LC .....(by C.p.c.t)

AM = CB....(by C.p.c.t ).....(1)...(B)

In ∆ ALC & ALB

$\angle {ALC} = \angle{MLB}$ ...( VOA )

$AL \: = \: LB$ ...( L is mid point of AB )

$\angle{CAL} = \angle{ LBM }$ ..(alt. int. angle)

Triangles are congruent by ASA property

AC = MB ...(by c.p.c.t)..(2)....(C)

$\angle{ACL} = \angle{LMB}$ ...(by c.p.c.t)

But these angles are alt. int. angles.

Alternate Interior Angles Theorem: If two lines are cut by a transversal and the alternate interior angles are congruent, then the lines are parallel.

Means AC || MB (by converse of alt. int. angle theorem ) ......(D)

By (A) , (B), (C) and (D)

ACBM is a parallelogram.