ABCD is a trapezium with $$AB\parallel DC$$. A line parallel to AC intersect AB at point M and BC at point N. Then prove area of $$\triangle ADM$$ = area of $$\triangle ACN$$.

Hint: So to understand it in better way we have to draw the diagram,

Here it is given that $$AB\parallel DC$$ and $$AC\parallel MN$$. We have to prove that the area of $$\triangle ADM$$ = area of $$\triangle ACN$$. So to find the solution we need to know a theorem, which states that “Two triangles on the same base (or equal bases) and between the same parallels are equal in area”.
By using the above theorem we are going to find the solution.

Complete step-by-step answer:
Given a trapezium ABCD where $$AB\parallel DC$$ and $$AC\parallel MN$$.
Now considering the triangles $$\triangle ACM$$ and $$\triangle ACN$$, lies on the same base AC and are between parallel lines AC and MN.
Therefore by the above theorem we can say that their area is equal,
i.e, Area of $$\triangle ACM$$ = Area of $$\triangle ACN$$..........(1)

Now for triangles $$\triangle ACM$$ and $$\triangle ADM$$,
They lie on the same base AM and are between parallel lines AM and DC. (Since $$AB\parallel DC$$)
Therefore by the above mentioned theorem we can write,
Area of $$\triangle ADM$$ = Area of $$\triangle ACM$$........(2)

Now from equation (1) and (2) we can write,
Area of $$\triangle ADM$$ = Area of $$\triangle ACN$$
Hence proved.

Note: So you might be thinking how the area of two triangle on the same base and between the same parallels are equal, so let us try to prove it,
As we know that the area of a triangle $$A=\dfrac{1}{2} \times \text{Base} \times \text{Height}$$