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: Understanding the Question:
This is a challenging geometry problem from the chapter "Triangles".
We are given a triangle $ABC$ with median $AD$.
A point $X$ lies on $AD$ dividing it in the ratio $AX : XD = 2 : 3$.
The line $BX$ is extended to intersect side $AC$ at point $Y$.
We need to prove that $BX$ is 4 times $XY$.

Step 2: Key Formula or Approach:
We will use a standard geometric construction to apply the Basic Proportionality Theorem (Thales' Theorem):
1. Draw a line $DP$ parallel to $BY$ (and thus parallel to $XY$) meeting $AC$ at point $P$.
2. Apply the Basic Proportionality Theorem in $\Delta ADP$ and $\Delta BCY$ to relate the segments.

Step 3: Detailed Explanation:

• Construction:
Draw a line $DP \parallel BY$ through $D$, meeting $AC$ at point $P$.
Since $D$ lies on $BC$ and $P$ lies on $AC$, $DP \parallel BY$ implies $DP \parallel XY$.

• In Triangle ADY:
Since $XY \parallel DP$, we can apply Thales' Theorem:
\[ \frac{AX}{XD} = \frac{AY}{YP} \] We are given that $\frac{AX}{XD} = \frac{2}{3}$. Thus:
\[ \frac{AY}{YP} = \frac{2}{3} \implies YP = \frac{3}{2} AY \quad \text{--- (Equation 1)} \]

• In Triangle BCY:
Since $DP \parallel BY$, we can apply Thales' Theorem:
\[ \frac{CP}{PY} = \frac{CD}{DB} \] Since $AD$ is a median, $D$ is the midpoint of $BC$, which means $CD = DB$:
\[ \frac{CP}{PY} = 1 \implies CP = PY \quad \text{--- (Equation 2)} \]

• Using Similarity of Triangles:
- In $\Delta ADP$, since $XY \parallel DP$:
$\Delta AXY \sim \Delta ADP$ by AA similarity.
The ratio of corresponding sides is:
\[ \frac{XY}{DP} = \frac{AX}{AD} = \frac{2}{2 + 3} = \frac{2}{5} \implies DP = \frac{5}{2} XY \quad \text{--- (Equation 3)} \] - In $\Delta CBY$, since $DP \parallel BY$:
$\Delta CDP \sim \Delta CBY$ by AA similarity.
The ratio of corresponding sides is:
\[ \frac{DP}{BY} = \frac{CP}{CY} \] Since $CP = PY$, we have $CY = CP + PY = 2 CP$. Thus:
\[ \frac{DP}{BY} = \frac{CP}{2 CP} = \frac{1}{2} \implies BY = 2 DP \quad \text{--- (Equation 4)} \]

• Combining Equations 3 and 4:
Substitute $DP = \frac{5}{2} XY$ into Equation 4:
\[ BY = 2 \left( \frac{5}{2} XY \right) = 5 XY \]

• Finding BX:
From the line segment $BY$, we know:
\[ BY = BX + XY \] Therefore:
\[ BX + XY = 5 XY \] \[ BX = 5 XY - XY = 4 XY \]

Step 4: Final Answer:
Hence Proved.