# X and Y are points on the side LN of the triangle LMN such that LX = XY = YN. Through X, a line is drawn parallel to LM to meet MN at Z (See Fig. 9.12). Prove that ar (LZY) = ar (MZYX)

**Solution:**

From the details given

LX=XY=YN

XZ II LM

We have to prove that

ar (LZY) = ar (MZYX)

From the figure

ar(LZX)+(XZY)=ar(LZY) …. (i)

ar(MXZ)+ar(XZY)=ar(MZYX) …. (ii)

The triangles LZX and MXZ are on the same base XZ and between the same parallels LM and XZ

So we get

ar(LZX)=ar(MXZ)

By adding both the equations

ar(LZY)=ar(MZYX)

Therefore, it is proven that ar(LZY) = ar(MZYX)

**✦ Try This:** Diagonals AC and BD of a quadrilateral ABCD intersect each other at P show that ar.(△APB) × ar.(△CPD) = ar.(△APD) × ar.(△BPC)

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

**NCERT Exemplar Class 9 Maths Exercise 9.3 Problem 2**

## X and Y are points on the side LN of the triangle LMN such that LX = XY = YN. Through X, a line is drawn parallel to LM to meet MN at Z (See Fig. 9.12). Prove that ar (LZY) = ar (MZYX)

**Summary:**

X and Y are points on the side LN of the triangle LMN such that LX = XY = YN. Through X, a line is drawn parallel to LM to meet MN at Z (See Fig. 9.12). It is proven that ar (LZY) = ar (MZYX)

**☛ Related Questions:**

- The area of the parallelogram ABCD is 90 cm² (see Fig.9.13). Find (i) ar (ABEF), (ii) ar (ABD), (iii . . . .
- In ∆ ABC, D is the mid-point of AB and P is any point on BC. If CQ || PD meets AB in Q (Fig. 9.14), . . . .
- ABCD is a square. E and F are respectively the midpoints of BC and CD. If R is the mid-point of EF ( . . . .

visual curriculum