# In Fig. 9.26, X and Y are the mid-points of AC and AB respectively, QP || BC and CYQ and BXP are straight lines. Prove that ar (ABP) = ar (ACQ)

**Solution:**

Given, CYQ and BXP are straight lines

X and Y are the midpoints of AC and AB

QP | | BC

We have to prove that ar(ABP) = ar(ACQ)

Since X is the midpoint of AC

AX = CX

Since Y is the midpoint of AB

AY = BY

Therefore, XY || BC

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

Triangles BYC and BXC are on the same base BC and between the same parallel lines XY and BC.

So, ar(BYC) = ar(BXC) ----------------------- (1)

Subtracting ar(BOC) on both sides,

ar(BYC) - ar(BOC) = ar(BXC) - ar(BOC)

So, ar(BOY) = ar(COX) ----------------------- (2)

On adding ar(XOY) on both sides,

ar(BOY) + ar(XOY) = ar(COX) + ar(XOY)

So, ar(BXY) = ar(CXY) ----------------------- (3)

We know that the quadrilaterals on the same base and between the same parallel are equal in areas.

We observe that quadrilaterals XYAP and YXAQ are on the same base XY and between the same parallels XY and PQ.

ar (XYAP) = ar (YXAQ) ---------------------- (4)

On adding (3) and (4),

ar (ΔBYX) + ar (XYAP) = ar (ΔCXY) + ar (YXAQ)

From the figure,

ar (ΔCXY) + ar (YXAQ) = ar(ACQ)

ar (ΔBYX) + ar (XYAP) = ar(ABP)

Therefore, ar(ABP) = ar(ACQ)

**✦ Try This: **In the adjoining figure, DE ∣∣ BC. Prove that ar(△OCE) = ar(△OBD)

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

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

## In Fig. 9.26, X and Y are the mid-points of AC and AB respectively, QP || BC and CYQ and BXP are straight lines. Prove that ar (ABP) = ar (ACQ)

**Summary:**

In Fig. 9.26, X and Y are the mid-points of AC and AB respectively, QP || BC, CYQ and BXP are straight lines. It is proven that ar (ABP) = ar (ACQ)

**☛ Related Questions:**

- In ∆ ABC, if L and M are the points on AB and AC, respectively such that LM || BC. Prove that ar (LO . . . .
- In Fig. 9.25, ABCDE is any pentagon. BP drawn parallel to AC meets DC produced at P and EQ drawn par . . . .
- If the medians of a ∆ ABC intersect at G, show that ar(AGB) = ar(AGC) = ar(BGC) = 1/3 ar(ABC)

visual curriculum