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Let $A B C$ be a triangle with $A B = A C$ .If $D$ is the midpoints of $B C$ . $E$ the foot of perpendicular drawn from $D$ on $A C$ and $F$ the midpoint of $D E$ ,then prove that $A F$ is perpendicular to $B E$ by vector method

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Solution

Let $- - \to A B = \to b$ and $- - \to A C = \to c$ where $∣ ∣ ∣ \to b ∣ ∣ ∣ = ∣ ∣ \to c ∣ ∣$
Let $- - \to A E = α \to c$
But $- - \to A D = \frac{\to b + \to c}{2}$
$- - \to D E ⊥ - - \to A C \Rightarrow ⎛ ⎝ α \to c - \frac{\to b + \to c}{2} ⎞ ⎠ . \to c = 0$
$\Rightarrow α {∣ ∣ \to c ∣ ∣}^{2} - \frac{\to b + \to c}{2} . \to c = 0$
$\Rightarrow \to b . \to c = (2 α - 1) {∣ ∣ \to c ∣ ∣}^{2}$ ...... $(1)$
$- - \to A F = \frac{1}{2} ⎛ ⎝ \frac{\to b + \to c}{2} + α \to c ⎞ ⎠$
$= \frac{1}{4} (\to b + (2 α + 1) \to c)$
$- - \to B E = α \to c - \to b$
$∴ - - \to A F . - - \to B E = \frac{1}{4} [(2 α^{2} + α - 1) {∣ ∣ \to c ∣ ∣}^{2} - (1 + α) \to b . \to c]$ ...... $(2)$
Since $∣ ∣ ∣ \to b ∣ ∣ ∣ = ∣ ∣ \to c ∣ ∣$
Now using $(1)$ in $(2)$ we find that $- - \to A F . - - \to B E = 0$
$∴ - - \to A F ⊥ - - \to B E$

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