Let ABC be a triangle with AB = AC. If...

MathematicsAngle Between Two LinesJEE Advanced 1989Moderate

Let $A B C$ be a triangle with $A B = A C$ . If $D$ is the midpoint of $B C$ , $E$ is the foot of the perpendicular drawn from $D$ to $A C$ and $F$ the mid-point of $D E$ , prove that $A F$ is perpendicular to $B E$ .

Visualized Solution (Hindi)

Coordinate Setup

Let $D$ be the origin $(0, 0)$ .
Let $B C$ lie on the x-axis.
Coordinates: $B (- a, 0)$ , $C (a, 0)$ , $A (0, h)$ .

Equation of Line .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A C

Points: $A (0, h)$ and $C (a, 0)$
Equation of $A C$ : $y - 0 = \frac{h - 0}{0 - a} (x - a)$
Simplified: $h x + a y - ah = 0$

Finding Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } E

$E$ is the foot of perpendicular from $D (0, 0)$ to $h x + a y - ah = 0$ .
Using formula: $\frac{x - 0}{h} = \frac{y - 0}{a} = - \frac{h ( 0 ) + a ( 0 ) - ah}{h ^{2} + a ^{2}}$
$E = (\frac{a h ^{2}}{a ^{2} + h ^{2}}, \frac{a ^{2} h}{a ^{2} + h ^{2}})$

Finding Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } F

$F$ is the midpoint of $D E$ , where $D (0, 0)$ and $E (x_{E}, y_{E})$ .
$F = (\frac{x _{E}}{2}, \frac{y _{E}}{2})$
$F = (\frac{a h ^{2}}{2 ( a ^{2} + h ^{2} )}, \frac{a ^{2} h}{2 ( a ^{2} + h ^{2} )})$

Slope of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A F

Slope $m_{1}$ of $A F$ where $A (0, h)$ and $F (x_{F}, y_{F})$ .
$m_{1} = \frac{y _{F} - h}{x _{F} - 0} = \frac{\frac{a ^{2} h}{2 ( a ^{2} + h ^{2} )} - h}{\frac{a h ^{2}}{2 ( a ^{2} + h ^{2} )}}$
$m_{1} = \frac{a ^{2} h - 2 h ( a ^{2} + h ^{2} )}{a h ^{2}} = - \frac{a ^{2} + 2 h ^{2}}{ah}$

Slope of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B E

Slope $m_{2}$ of $B E$ where $B (- a, 0)$ and $E (x_{E}, y_{E})$ .
$m_{2} = \frac{y _{E} - 0}{x _{E} - ( - a )} = \frac{\frac{a ^{2} h}{a ^{2} + h ^{2}}}{\frac{a h ^{2}}{a ^{2} + h ^{2}} + a}$
$m_{2} = \frac{a ^{2} h}{a h ^{2} + a ( a ^{2} + h ^{2} )} = \frac{ah}{a ^{2} + 2 h ^{2}}$

Final Proof: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } m_{1} \cdot m_{2} = - 1

Product of slopes: $m_{1} \cdot m_{2}$
$(- \frac{a ^{2} + 2 h ^{2}}{ah}) \cdot (\frac{ah}{a ^{2} + 2 h ^{2}}) = - 1$
Since $m_{1} \cdot m_{2} = - 1$ , $A F ⊥ B E$ .

Summary and Takeaway

Key Takeaway: Coordinate geometry simplifies geometric proofs through algebraic relations.
Symmetry: Placing the midpoint at the origin is a powerful strategy.
Next Challenge: Try proving this using vectors or pure Euclidean geometry.

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