Question

ABCD is a trapezium with AB II DC. A line parallel to AC intersects AB at X and BC at Y. Prove that ar (ADX) = ar (ACY). [Hint : Join CX.].

Answer 1

Step-by-step explanation:

In the question,

ABCD is a trapezium.

AB || CD. Also, XY is || AC.

So,

Let us join CX.

Now, we know the area of triangle is given by,

$A=\frac{1}{2}\times base\times height$

So,

In triangle ADX,

$A=\frac{1}{2}\times AX \times h$

where h is the height of the trapezium.

Also,

In triangle ACX,

$A=\frac{1}{2}\times AX \times h$

So,

Area of ADX = Area of ACX

Now,

In triangle ABC as XY is parallel to AC.

So,

Area of triangle ACY = Area of triangle ACX

(Area of the triangles having the same base between the parallel lines are equal.)

So,

We can say that,

Area (ADX) = Area (ACY)

Hence, Proved.