If two triangles are on the same base and between the same parallel lines then they will be

$$\triangle$$ ABD and $$\triangle$$ ABC are on same base AB.

AB is parallel to CD, => ABCD is parallelogram. CE is height of the parallelogram

ar ($$\triangle$$ ABD) = $$\frac{1}{2} \times$$ base $$\times$$ height

=> $$ar (\triangle ABD) = \frac{1}{2} \times AB \times CE$$ ------------- (i)

Again, ar ($$\triangle$$ ABC) = $$\frac{1}{2} \times$$ base $$\times$$ height

=> $$ar (\triangle ABC) = \frac{1}{2} \times AB \times CE$$ ------------- (ii)

Equating (i) & (ii), we get :

=> $$ar (\triangle ABD) = ar (\triangle ABC)$$

=> Ans - (C)