Question

Answer quickly ANd brief

Answer 1

Answer:

In triangle AMD and triangle CNB

Angle CBN=AngleADM(Alternate angles

AM=CN(Given)

AngleAMD=AngleCNB

AMD =CNB(BY ASA CONGRUENCE)

Answer 2

Answer:

YES, ΔAMD=ΔCNB

Step-by-step explanation:

Hii mate,

AM and CN are perpendicular to the diagonal BD of a parallelogram ABCD

In parallelogram ABCD ,

AB = CD

BC = AD

BD is a diagonal .

In  ΔABD & ΔCDB ,

AB = CD

AD = CB

BD = BD ( common)

ΔABD ≅ ΔCDB

So, ∠ABD=∠BDC

Finally,

In ΔAMD & ΔCNB,

AB=CD (Opposite sides of a parallelogram are equal)

∠ABD=∠BDC (Proved above)

∠AMB=∠CND=90°

So, using AAS congruency theorem,

here 2 angles and a non-included sides are equal

Hence, ΔAMD≅ΔCNB

or in other words,

ΔAMD=ΔCNB

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