X and Y are points on the side LN of the ALMN such that LX = XY = YN.

1.	X and Y are points on the side LN of the ALMN such that LX = XY = YN. Through X, a line is drawn parallel to LM to meet MN at Z (see figure). Prove that ar (∆LZY) = ar (quadrilateral MZYX).
Answer» Given: X and Y are the points on side LN of ∆LMN, such that LX = XY = YN and LM \|\| XZ. To prove: ar (∆LZY) = ar (quad. MZYX) Proof: Since ∆XMZ and ∆ZLX are on the same base XZ and between same parallel LM and XZ. Then, ar (∆XMZ) = ar (∆ZLX) …(i) On adding ar (∆XYZ) on both sides of equation (i), we get ar (∆XMZ) + ar (∆XYZ) = ar (∆ZLX) + ar (∆XYZ) ⇒ ar (quad. MZYX) = ar (∆LZY) Hence proved.

X and Y are points on the side LN of the ALMN such that LX = XY = YN. Through X, a line is drawn parallel to LM to meet MN at Z (see figure). Prove that ar (∆LZY) = ar (quadrilateral MZYX).

Answer»

Given: X and Y are the points on side LN of ∆LMN, such that LX = XY = YN and LM || XZ.

To prove: ar (∆LZY) = ar (quad. MZYX)

Proof: Since ∆XMZ and ∆ZLX are on the same base XZ and between same parallel LM and XZ.

Then, ar (∆XMZ) = ar (∆ZLX) …(i)

On adding ar (∆XYZ) on both sides of equation (i), we get

ar (∆XMZ) + ar (∆XYZ) = ar (∆ZLX) + ar (∆XYZ)

⇒ ar (quad. MZYX) = ar (∆LZY)

Hence proved.