Question

Help Me!

Answer 1

Answer:

Given:

• DC ║ AB
• CM = MB as M is midpoint of BC

i) Since DN and BC are transversals, we have:

• ∠DCM ≅ ∠NBM and
• ∠CDM ≅ ∠BNM as alternate interior angles

As two angles and one side is congruent, the triangles are also congruent:

• ΔDCM ≅ ΔNBM (according to AAC postulate)

So their areas are same.

ii)

The quadrilateral has area of:

• A(ADCB) = A(ADMB) + A(DCM)

And the triangle has area of:

• A(ADN) = A(ADMB) + A(NBM)

Since the areas of triangles DCM and NBM are same, the quadrilateral ADCB has same area as triangle ADN.

Answer 2

Answer:

I think I have proved it before you asked this question before also.

Step-by-step explanation:

SEE the image for solution.

HOPE it helps

Have a great day