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 intersecting line from the midpoint of the base is a classic, highly effective technique for solving median-based ratio questions in triangles.

Answer 1

Step 1: Set up coordinates instead of drawing an auxiliary construction line.
Place $B$ at the origin, $(0,0)$, and $C$ at $(2a,0)$ for some $a>0$, so that the midpoint $D$ (the foot of the median) is at $(a,0)$. Let $A$ be an arbitrary point $(p,q)$ with $q \neq 0$.

Step 2: Find $X$ on $AD$ using the given ratio.
Since $AX:XD = 2:3$, point $X$ divides $A$ to $D$ in the ratio $2:3$, so by the section formula:
\[ X = \left(\frac{3p+2a}{5},\ \frac{3q}{5}\right) \]

Step 3: Write the line through $B$ and $X$, and the line through $A$ and $C$, using one parameter each.
Any point on ray $BX$ can be written as $t\cdot X$ for a real number $t \geq 0$ (since $B$ is the origin).
Any point on segment $AC$ can be written as $A + s(C-A)$ for $0 \leq s \leq 1$.
Point $Y$ is where these two lines meet, so we need $t\cdot X = A + s(C-A)$.

Step 4: Solve for $t$, which directly tells us the ratio $BY:BX$.
Matching the $y$-coordinates: $t\cdot\dfrac{3q}{5} = q(1-s)$, so $s = 1 - \dfrac{3t}{5}$.
Matching the $x$-coordinates and substituting this $s$, then simplifying (the $p$ terms cancel out), gives:
\[ 10a = 8at \implies t = \frac{5}{4} \]

Step 5: Interpret $t=\dfrac{5}{4}$ geometrically.
Since $X$ itself sits at parameter $t=1$ along ray $BX$ (because $X = 1\cdot X$), and $Y$ sits at parameter $t=\dfrac{5}{4}$ along the same ray, the point $Y$ is $\dfrac{5}{4}$ times as far from $B$ as $X$ is. So:
\[ BY = \frac{5}{4}BX \]
\[ XY = BY - BX = \frac{5}{4}BX - BX = \frac{1}{4}BX \]
\[ BX = 4\,XY \]

Final Answer:
Using coordinates confirms the same relation.
\[ \boxed{BX = 4\,XY} \]