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

Then, equation of $AC$ is
$\frac{x}{a}+\frac{y}{h}=\mathrm{1.....}$(i)
and equation of $DE\perp AC$ and passing through origin is
$\frac{x}{h}-\frac{y}{a}=0\Rightarrow x=\frac{hy}{a}....$(ii)
On solving, Eqs. (i) and (ii), we get the coordinates of point $E$ as follows
$\frac{hy}{a^2}+\frac{y}{h}=1$
$\Rightarrow y=\frac{a^{2}h}{a^2+h^2}$
$\therefore$ 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 $DE$.
$\therefore$ 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 $AF$,
$m_1=\frac{h-\frac{a^{2}h}{2\left(a^2+h^2\right)}}{0-\frac{a{h}^2}{2\left(a^2+h^2\right)}}=\frac{2h\left(a^2+h^2\right)-a^{2}h}{-a{h}^2}$
$\Rightarrow m_1=\frac{-\left(a^2+2h^2\right)}{ah}....$(iii)
and slope of $BE,m_2=\frac{\frac{a^{2}h}{a^2+h^2}-0}{\frac{a{h}^2}{a^2+h^2}+a}$
$=\frac{a^{2}h}{a{h}^2+a^3+a{h}^2}$
$\Rightarrow m_2=\frac{ah}{a^2+2h^2}......$(iv)
From Eqs. (iii) and (iv),
$m_1{m}_2=-1\Rightarrow AF\perp BE$