# Q: 13      ABCD is a trapezium with $\small AB\parallel DC$. A line parallel to AC intersects AB at X and BC at Y. Prove that $\small ar(ADX)=ar(ACY)$. [Hint: Join CX.]

We have a trapezium ABCD, AB || CD
XY ||AC meets AB at X and BC at Y. Join XC

Since $\Delta$ADX and $\Delta$ACX lie on the same base CD and between same parallels AX and CD
Therefore, ar($\Delta$ADX) = ar($\Delta$ACX)..........(i)
Similarly ar($\Delta$ACX) = ar($\Delta$ACY).............(ii) [common base AC and AC || XY]
From eq (i) and eq (ii), we get

ar($\Delta$ADX) = ar ($\Delta$ACY)

Hence proved.

## Related Chapters

### Preparation Products

##### JEE Main Rank Booster 2021

This course will help student to be better prepared and study in the right direction for JEE Main..

₹ 13999/- ₹ 9999/-
##### Rank Booster NEET 2021

This course will help student to be better prepared and study in the right direction for NEET..

₹ 13999/- ₹ 9999/-
##### Knockout JEE Main April 2021 (Easy Installments)

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 4999/-
##### Knockout NEET May 2021

An exhaustive E-learning program for the complete preparation of NEET..

₹ 22999/- ₹ 14999/-