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)

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