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: Understanding the Question:
In a triangle \(\Delta ABC\), \(AD\) is a median, which means \(D\) is the midpoint of the base \(BC\).
\(X\) is a point on \(AD\) with the division ratio \(AX : XD = 2 : 3\).
The line segment \(BX\) is extended to meet the side \(AC\) at point \(Y\).
We need to prove that the length segment \(BX = 4 XY\).

Step 2: Key Formula or Approach:
To prove this relation, we can use a geometric construction:
1. Draw a line \(DK\) parallel to \(BY\) (or \(XY\)) which intersects \(AC\) at point \(K\).
2. Apply the Basic Proportionality Theorem (BPT) in \(\Delta ADK\) and \(\Delta BCY\) to establish ratios between the segments.

Step 3: Detailed Explanation:
1. Construction: Draw a line segment \(DK\) from point \(D\) to side \(AC\) such that \(DK \parallel BY\).
2. Consider \(\Delta ADK\):
- Since \(DK \parallel BY\), it is also parallel to \(XY\) (as \(X, Y, B\) are collinear).
- By the Basic Proportionality Theorem (Thales's Theorem):
\[ \frac{AY}{YK} = \frac{AX}{XD} \]
- We are given \(AX : XD = 2 : 3\), so:
\[ \frac{AY}{YK} = \frac{2}{3} \implies AY = \frac{2}{3}YK \quad \text{--- (Equation 1)} \]
3. Consider \(\Delta BCY\):
- Since \(D\) is the midpoint of \(BC\) (as \(AD\) is a median), \(BD = DC\).
- By construction, \(DK \parallel BY\).
- By the Converse of the Midpoint Theorem, since \(DK\) is parallel to \(BY\) and passes through the midpoint \(D\) of \(BC\), it must also bisect \(CY\).
- Therefore, \(K\) is the midpoint of \(CY\), meaning:
\[ YK = KC \quad \text{--- (Equation 2)} \]
4. Now, find the relationship between the lengths \(XY\) and \(DK\) using similar triangles \(\Delta AYX \sim \Delta AKD\):
\[ \frac{XY}{DK} = \frac{AX}{AD} \]
- Since \(AD = AX + XD\), we have \(\frac{AX}{AD} = \frac{2}{2+3} = \frac{2}{5}\).
- Therefore:
\[ XY = \frac{2}{5}DK \quad \text{--- (Equation 3)} \]
5. Using the Midpoint Theorem in \(\Delta BCY\), the segment \(DK\) connecting the midpoints of \(BC\) and \(CY\) is half the length of \(BY\):
\[ DK = \frac{1}{2}BY \]
6. Substitute this into Equation 3:
\[ XY = \frac{2}{5}\left(\frac{1}{2}BY\right) = \frac{1}{5}BY \]
7. Since \(BY = BX + XY\):
\[ XY = \frac{1}{5}(BX + XY) \]
\[ 5XY = BX + XY \]
\[ BX = 4XY \]
8. Hence, the relation is successfully proved.

Step 4: Final Answer:
The relation \(BX = 4 XY\) is proved.