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  • 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.). Prove that ar (LZY) = ar (MZYX)

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.). Prove that ar (LZY) = ar (MZYX)

 

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Solution.

Given: \triangle LMN with X and Y points on the side LN
LX = XY = YN and XZ \parallel LM
To prove: ar(\triangle LZY) = ar(MZYX)
Proof:
Since XZ \parallel LM and \triangle XMZ and \triangle XLZ are on the same base XZ and between the same parallel lines LM and XZ. Then,
ar(\triangle XMZ)=ar (\triangle XLZ)                        …(1)
Adding ar(\triangle XYZ)  both sides of eq. (1), we get
ar(\triangle XMZ) + ar(\triangle XYZ) = ar(\triangle XLZ) + ar(\triangle XYZ)
\Rightarrow ar(XMZY) = ar(LZY)
This can be rearranged as:
ar (LZY) = ar (MZYX)
Hence proved.

 

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