In △ 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.…

Sol. Draw DZ $\|\|$ BY. (1/2 Mark)
In $\triangle CBY$, DZ $\|\|$ BY
$\therefore \frac{CD}{DB} = \frac{CZ}{ZY} = 1$
Therefore, DZ $= \frac{1}{2}$ BY --- (i) (1/2 Mark)
In $\triangle ADZ$, DZ $\|\|$ XY
So, $\triangle AXY \sim \triangle ADZ$
$\therefore \frac{AX}{AD} = \frac{XY}{DZ}$ (1 Mark)
$\Rightarrow \frac{2}{5} = \frac{XY}{DZ}$ or DZ $= \frac{5}{2}$ XY --- (ii) (1/2 Mark)
Using (i) and (ii),
BY $= 5$ XY (1 Mark)
Therefore BX = BY – XY $= 4$ XY (1/2 Mark)
Alternate solution:
Draw DZ $\|\|$ BY. (1/2 Mark)
$\triangle AXY \sim \triangle ADZ$ (1/2 Mark)
$\therefore \frac{AX}{AD} = \frac{XY}{DZ} \Rightarrow \frac{2}{2+3} = \frac{XY}{DZ}$
$\Rightarrow 2$ DZ $= 5$ XY --- (i) (1 Mark)
Now, $\triangle CDZ \sim \triangle CBY$
$\therefore \frac{CD}{CB} = \frac{DZ}{BY}$ (1/2 Mark)
But CD $= \frac{1}{2}$ BC
So, $\frac{DZ}{BY} = \frac{1}{2} \Rightarrow BY = 2$ DZ --- (ii) (1 Mark)
From (i) and (ii), we get
BY $= 5$ XY (1 Mark)
$\Rightarrow (BX + XY) = 5$ XY
$\Rightarrow BX = 4$ XY (1/2 Mark)