`A B C D` is a trapezium with `A B D Cdot` A line parallel to `A C` intersects `A B` at `X` and `B C` at `Ydot` Prove that `a r( A D X)=a r( A C Y)dot`

Explanation

Given: `ABCD` is a trapezium with `AB∥DC` and `XY∥AC`.
Join `YA`, `XC` and `XD`.
`triangleACX` and `triangleACY` have same base `AC` and are between same parallels `AC` and `XY`.
So, `ar(△ACX)=ar(△ACY)` ......(i)
`triangleACX` and `triangleADX` have same base `AX` and are between same parallels `AB` and `DC`.
So, `ar(△ACX)=ar(△ADX)` ......(ii)
From (i) and (ii), we get
`ar(△ADX)=ar(△ACY)`
Hence Proved.