Question

Let $ABC$ be a triangle with $AB= AC$. If $D$ is mid point of $BC$, the foot of the perpendicular drawn from $D$ to $AC$ and $F$ the mid-point of $DE$. Prove that $AF$ is perpendicular to $BE$.

Answer 1

Let $B C$ be taken as $X$-axis with origin at $D$, the mid-point of $B C$ and $D A$ will be $Y$-axis.
Given, $A B=A C$
Let $B C=2 a$, then the coordinates of $B$ and $C$ are $(-a, 0)$ and $(a, 0)$ let $A(0, h)$.

Then, equation of $A C$ is
$\frac{x}{a}+\frac{y}{h}=1.....$(i)
and equation of $D E \perp A C$ and passing through origin is
$\frac{x}{h}-\frac{y}{a}=0 \Rightarrow x=\frac{h y}{a}....$(ii)
On solving, Eqs. (i) and (ii), we get the coordinates of point $E$ as follows
$\frac{h y}{a^{2}}+\frac{y}{h}=1$
$\Rightarrow y=\frac{a^{2} h}{a^{2}+h^{2}}$
$\therefore $ Coordinate of $ E=\left(\frac{a h^{2}}{a^{2}+h^{2}}, \frac{a^{2} h}{a^{2}+h^{2}}\right)$
Since, $F$ is mid-point of $D E$.
$\therefore$ Coordinate of $F\left[\frac{a h^{2}}{2\left(a^{2}+h^{2}\right)}, \frac{a^{2} h}{2\left(a^{2}+h^{2}\right)}\right]$
$\therefore$ Slope of $A F$,
$m_{1}=\frac{h-\frac{a^{2} h}{2\left(a^{2}+h^{2}\right)}}{0-\frac{a h^{2}}{2\left(a^{2}+h^{2}\right)}}=\frac{2 h\left(a^{2}+h^{2}\right)-a^{2} h}{-a h^{2}}$
$\Rightarrow m_{1}=\frac{-\left(a^{2}+2 h^{2}\right)}{a h}....$(iii)
and slope of $BE, m_{2}=\frac{\frac{a^{2} h}{a^{2}+h^{2}}-0}{\frac{a h^{2}}{a^{2}+h^{2}}+a}$
$=\frac{a^{2} h}{a h^{2}+a^{3}+a h^{2}}$
$\Rightarrow m_{2}=\frac{a h}{a^{2}+2 h^{2}}......$(iv)
From Eqs. (iii) and (iv),
$m_{1} m_{2}=-1 \Rightarrow A F \perp B E$