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 parallel line through the midpoint of a side is a very powerful auxiliary construction for triangle proofs.
It allows you to use both Thales' Theorem and the Midpoint Theorem in different sub-triangles to connect ratios.

Answer 1

Step 1: Assign masses at B and C using the fact that D is the midpoint of BC.
Since $AD$ is a median, $D$ is the midpoint of $BC$. In mass point geometry, equal masses at the two endpoints of a segment balance at its midpoint, so assign:
\[ \text{mass}(B) = 1, \qquad \text{mass}(C) = 1 \]
The mass at $D$, which balances $B$ and $C$, is the sum:
\[ \text{mass}(D) = \text{mass}(B) + \text{mass}(C) = 1 + 1 = 2 \]

Step 2: Use the given ratio AX:XD to find the mass at A.
In mass point geometry, a point divides a segment in the ratio inverse to the masses at its ends, so for $X$ on $AD$ with $AX:XD = 2:3$:
\[ \frac{\text{mass}(A)}{\text{mass}(D)} = \frac{XD}{AX} = \frac{3}{2} \]
\[ \text{mass}(A) = \frac{3}{2} \times \text{mass}(D) = \frac{3}{2} \times 2 = 3 \]

Step 3: Find the mass at Y using the cevian AC, since Y lies on AC.
Since $Y$ lies on side $AC$, and we now know $\text{mass}(A) = 3$ and $\text{mass}(C) = 1$, the mass at $Y$ (balancing $A$ and $C$) is:
\[ \text{mass}(Y) = \text{mass}(A) + \text{mass}(C) = 3 + 1 = 4 \]

Step 4: Use the masses at B and Y to find the ratio BX:XY along line BY.
Point $X$ also lies on the cevian $BY$ (since $B$, $X$, $Y$ are collinear by construction, as $BX$ extended meets $AC$ at $Y$). Along this line, $X$ balances $B$ (mass $1$) and $Y$ (mass $4$), so:
\[ \frac{BX}{XY} = \frac{\text{mass}(Y)}{\text{mass}(B)} = \frac{4}{1} \]

Step 5: Write this ratio as the required equation.
\[ BX = 4\,XY \]

Final Answer:
Hence Proved, $BX = 4XY$.