# ABCD is a trapezium with AB || DC. A line parallel to AC intersects AB at X and BC at Y. Prove that ar (ADX) = ar (ACY).

[Hint: Join CX.]

**Solution:**

Let's draw the given trapezium ABCD with AB || DC.

We observe that ΔADX and ΔACX are lying on the same base AX and are existing between the same parallels AB and DC.

According to Theorem 9.2: Two triangles on the same base (or equal bases) and between the same parallels are equal in area.

Therefore, ar (ΔADX) = ar (ΔACX) ... (1)

Similarly, ΔACY and ΔACX are lying on the same base AC and are existing between the same parallels AC and XY.

Therefore, Area (ΔACY) = Area (ΔACX) ... (2)

From Equations (1) and (2), we obtain

Area (ΔADX) = Area (ΔACY)

Henced proved.

**☛ Check: **Class 9 Maths NCERT Solutions Chapter 9

**Video Solution:**

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

Maths NCERT Solutions Class 9 Chapter 9 Exercise 9.3 Question 13

**Summary:**

If ABCD is a trapezium with AB || DC, a line parallel to AC intersects AB at X and BC at Y, then ar (ADX) = ar (ACY).

**☛ Related Questions:**

- In Fig.9.23, E is any point on median AD of a ∆ ABC. Show that ar (ABE) = ar (ACE).
- In a triangle ABC, E is the mid-point of median AD. Show that ar (BED) = 1/4 ar(ABC).
- Show that the diagonals of a parallelogram divide it into four triangles of equal area.
- In Fig. 9.24, ABC and ABD are two triangles on the same base AB. If line- segment CD is bisected by AB at O, show that ar(ABC) = ar (ABD).