In Δ ABC, AD is a median. X is a point on AD such that AX: XD =2: 3. BX is extended so that it intersects AC at Y. Prove that BX =4 XY.

Draw DZ || BY. In \(\displaystyle \triangle \mathrm{CBY}\), DZ \(\displaystyle |\mid \mathrm{BY}\) \[\therefore \frac{\mathrm{CD}}{\mathrm{DB}}=\frac{\mathrm{CZ}}{\mathrm{ZY}}=1 \] Therefore, \(\displaystyle \mathrm{DZ}=\frac{1}{2} \mathrm{BY}\) In \(\displaystyle \triangle \mathrm{ADZ}, \mathrm{DZ} \| \mathrm{XY}\) So, \(\displaystyle \Delta \mathrm{AXY} \sim \Delta \mathrm{ADZ}\) \[\therefore \frac{A X}{A D}=\frac{X Y}{D Z} \] Using (i) and (ii), \[\text { BY = } 5 \text { XY } \] Therefore BX \(\displaystyle =\mathrm{BY}-\mathrm{XY}=4 \mathrm{XY}\) Alternate solution: Draw DZ || BY. \[\Delta \mathrm{AXY} \sim \Delta \mathrm{ADZ} \] \(\displaystyle \therefore \frac{\mathrm{AX}}{\mathrm{AD}}=\frac{\mathrm{XY}}{\mathrm{DZ}} \Rightarrow \frac{2}{2+3}=\frac{\mathrm{XY}}{\mathrm{DZ}}\) \[\Rightarrow 2 \mathrm{DZ}=5 \mathrm{XY} \] Now, \(\displaystyle \Delta \mathrm{CDZ} \sim \Delta \mathrm{CBY}\) \[\therefore \frac{\mathrm{CD}}{\mathrm{CB}}=\frac{\mathrm{DZ}}{\mathrm{BY}} \] But \(\displaystyle \mathrm{CD}=\frac{1}{2} \mathrm{BC}\) \[\text { So, } \frac{D Z}{B Y}=\frac{1}{2} \Rightarrow B Y=2 D Z \] From (i) and (ii), we get \[\text { BY = } 5 \text { XY } \] \[\Rightarrow(\mathrm{BX}+\mathrm{XY})=5 \mathrm{XY} \] \[\Rightarrow \mathrm{BX}=4 \mathrm{XY} \]