# In Fig. 9.24, CD || AE and CY || BA. Prove that ar (CBX) = ar (AXY)

**Solution:**

Given, CD || AE

CY || BA

We have to prove that ar(CBX) = ar(AXY)

We know that the area of triangles on the same base and between the same parallel lines are equal.

Triangles ABY and ABC are on the same base BA and between the same parallel lines CY and BA.

So, ar(ABY) = ar(ABC) ------- (1)

Adding ar(ABX) on both sides,

ar(ABX) + ar(ABY) = ar(ABC) + ar(ABX)

From the figure,

ar(ABX) + ar(ABY) = ar(AXY)

So, ar(AXY) = ar(ABC) + ar(ABX)

From the figure,

ar(ABC) + ar(ABX) = ar(CBX)

Therefore, ar(CBX) = ar(AXY)

**✦ Try This: **In a figure, PQRS and EFRS are two parallelograms then prove that ar(MFR) = 1/2 ar(PQRS).

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

**NCERT Exemplar Class 9 Maths Exercise 9.4 Problem 4**

## In Fig. 9.24, CD || AE and CY || BA. Prove that ar (CBX) = ar (AXY)

**Summary:**

In Fig. 9.24, CD || AE and CY || BA. It is proven that ar (CBX) = ar (AXY)

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