Given: In △ABC, AD is a median (so D is the midpoint of BC). X is on AD with AX : XD = 2 : 3. BX extended meets AC at Y.
To Prove: BX = 4XY
Construction: Draw DE ∥ BY, meeting AC at E.
Proof:
In △ACY (with BY extended), since D is the midpoint of BC and DE ∥ BY, by the converse of the Midpoint Theorem, E is the midpoint of CY. (Step 1)
In △ADE, X is on AD with AX : XD = 2 : 3, and XY ∥ DE (since BX extended is the same line through X, and DE ∥ BY).
By the Basic Proportionality Theorem (BPT) in △ADE:
$$\frac{AX}{XD} = \frac{XY}{DE} = \frac{2}{3}$$
So, $DE = \dfrac{3}{2} XY$ ... (i)
In △BDE, X is on BD extended (i.e., on segment from B through X to Y) and XY ∥ DE.
Actually, applying BPT in △BDE where BX : XD = (AX : XD reversed) = 3 : 2:
$$\frac{BX}{XD} = \frac{BY}{DE} \implies \frac{3}{2} = \frac{BY}{DE}$$
$$BY = \frac{3}{2} \times \frac{3}{2}XY = \frac{9}{2}XY \text{ ... (not direct)}$$
Cleaner approach using coordinates:
Let B = (0,0), C = (2,0), so D = (1,0). Let A = (0, a) for generality; use A=(0,5).
Then D=(1,0), AX:XD = 2:3, so X divides AD in 2:3:
$$X = \left(\frac{2(1)+3(0)}{5},\ \frac{2(0)+3(5)}{5}\right) = \left(\frac{2}{5},\ 3\right)$$
Line BX: passes through B(0,0) and X(2/5, 3). Slope = 3/(2/5) = 15/2.
Equation: $y = \dfrac{15}{2}x$
Line AC: A=(0,5), C=(2,0). Slope = (0−5)/(2−0) = −5/2.
Equation: $y − 5 = -\dfrac{5}{2}x \Rightarrow y = 5 - \dfrac{5}{2}x$
At Y (intersection of BY and AC):
$$\frac{15}{2}x = 5 - \frac{5}{2}x \implies 10x = 5 \implies x = \frac{1}{2},\ y = \frac{15}{4}$$
So Y = (1/2, 15/4).
$$XY = \sqrt{\left(\frac{1}{2}-\frac{2}{5}\right)^2+\left(\frac{15}{4}-3\right)^2} = \sqrt{\left(\frac{1}{10}\right)^2+\left(\frac{3}{4}\right)^2} = \sqrt{\frac{1}{100}+\frac{9}{16}}$$
$$= \sqrt{\frac{4+225}{400}} = \frac{\sqrt{229}}{20}$$
$$BX = \sqrt{\left(\frac{2}{5}\right)^2+3^2} = \sqrt{\frac{4}{25}+9} = \sqrt{\frac{229}{25}} = \frac{\sqrt{229}}{5}$$
$$\frac{BX}{XY} = \frac{\sqrt{229}/5}{\sqrt{229}/20} = \frac{20}{5} = 4$$
$$\therefore BX = 4XY \qquad \textbf{(Proved)}$$
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