Question

In \(\Delta\) 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.

Hint:

Drawing a line parallel to the bisector or the transversal from the midpoint of the base is a standard construction trick in triangle geometry.
It instantly creates midpoints and similar triangles, converting a complex coordinate geometry problem into a simple ratio verification.

Answer 1

Step 1: Place the triangle on coordinate axes.
Since $D$ is the midpoint of $BC$ (as $AD$ is a median), it helps to put $B$ and $C$ on the x-axis so that $D$ sits exactly between them.
Let $B = (0, 0)$ and $C = (2, 0)$, so that the midpoint is $D = (1, 0)$.
Let $A = (0, a)$ for some value $a \neq 0$ (any position works for this proof, since only ratios of lengths are being compared, and those ratios do not change if the whole figure is stretched or moved).

Step 2: Find the coordinates of $X$ using the section formula.
$X$ lies on $AD$ such that $AX : XD = 2 : 3$. By the section formula, a point dividing the segment from $A(x_1,y_1)$ to $D(x_2,y_2)$ in the ratio $m:n$ has coordinates:
\[ \left( \frac{mx_2 + nx_1}{m+n}, \frac{my_2 + ny_1}{m+n} \right) \] Here $m = 2$, $n = 3$, $A = (0, a)$ and $D = (1, 0)$:
\[ X = \left( \frac{2(1) + 3(0)}{5}, \frac{2(0) + 3(a)}{5} \right) = \left( \frac{2}{5}, \frac{3a}{5} \right) \]
Step 3: Find the equation of line $BX$ and where it meets $AC$.
Line $BX$ passes through $B(0,0)$ and $X\left(\frac{2}{5}, \frac{3a}{5}\right)$. Its slope is:
\[ m_{BX} = \frac{3a/5 - 0}{2/5 - 0} = \frac{3a}{2} \] So the equation of line $BX$ is $y = \frac{3a}{2}x$.
Line $AC$ passes through $A(0, a)$ and $C(2, 0)$. Its slope is:
\[ m_{AC} = \frac{0 - a}{2 - 0} = -\frac{a}{2} \] So the equation of line $AC$ is $y = a - \frac{a}{2}x$.
$Y$ is the point where these two lines cross, so set the two expressions for $y$ equal:
\[ \frac{3a}{2}x = a - \frac{a}{2}x \] Divide every term by $a$ (allowed since $a \neq 0$):
\[ \frac{3}{2}x = 1 - \frac{1}{2}x \] \[ \frac{3}{2}x + \frac{1}{2}x = 1 \] \[ 2x = 1 \implies x = \frac{1}{2} \] Substitute back to find $y$:
\[ y = \frac{3a}{2} \times \frac{1}{2} = \frac{3a}{4} \] So $Y = \left(\frac{1}{2}, \frac{3a}{4}\right)$.

Step 4: Compute $BX$ and $XY$ using the distance formula.
\[ BX = \sqrt{\left(\frac{2}{5} - 0\right)^2 + \left(\frac{3a}{5} - 0\right)^2} = \sqrt{\frac{4}{25} + \frac{9a^2}{25}} = \frac{1}{5}\sqrt{4 + 9a^2} \] \[ XY = \sqrt{\left(\frac{1}{2} - \frac{2}{5}\right)^2 + \left(\frac{3a}{4} - \frac{3a}{5}\right)^2} = \sqrt{\left(\frac{1}{10}\right)^2 + \left(\frac{3a}{20}\right)^2} = \sqrt{\frac{1}{100} + \frac{9a^2}{400}} \] \[ XY = \sqrt{\frac{4 + 9a^2}{400}} = \frac{1}{20}\sqrt{4 + 9a^2} \]
Step 5: Compare the two lengths.
Divide $BX$ by $XY$:
\[ \frac{BX}{XY} = \frac{\frac{1}{5}\sqrt{4+9a^2}}{\frac{1}{20}\sqrt{4+9a^2}} = \frac{1/5}{1/20} = \frac{20}{5} = 4 \] So $BX = 4 \times XY$, which is exactly what needed to be shown, confirmed here through direct coordinates and distances rather than the parallel-line construction.

Final Answer:
Hence it is proved that $BX = 4XY$.
\[ \boxed{BX = 4XY} \]