Question

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

Answer 1

$\begin{array}{c}LX=XY=YN\left(given\right)\\ \because LM\parallel XZ\\ \Rightarrow DrawZL{\perp }_{r}LY\\ \Rightarrow In\Delta LZYXT\parallel YZ\\ \Rightarrow ar\Delta LZY=\frac{1}{2}\left(LY\right)XZL=\left(2XY\right)\times \frac{1}{2}\times \left(ZL\right)=\left(ZL\right)\times \left(XY\right)\\ \Rightarrow Inquadrilateral\left(MZYX\right)=\left(base\times height\right)\\ \Rightarrow ar\left(MZYX\right)=\left(XY\right)\times \left(ZL\right)\\ \Rightarrow Hence,ar\left(\Delta LZY\right)=arMZYXproved.\end{array}$