Question

Question 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 figure). Prove that ar (\(\Delta\) LZY) = ar (MZYX).

Answer 1

Given X and Y are points on the side LN such that;
LX = XY = YN and XZ || LM
To prove: ar ($\Delta$ LZY) = (MZYX)
Proof:
$\Delta$ XMZ and $\Delta$ XLZ are on the same base XZ and between the same parallel lines LM and XZ.
Then, ar ($\Delta$ XMZ) = ar ($\Delta$ XLZ) ...(i)
On adding ar ($\Delta$ XYZ) both sides of eq (i), we get,
ar ($\Delta$XMZ) + ar ($\Delta$XYZ) = ar ($\Delta$ XLZ) + ar ($\Delta$ XYZ)
$\Rightarrow$ ar (MZYX) = ar ($\Delta$ LZY)