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.