Question

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

• A. $BE$
• B. $BC$
• C. $DE$
• D. $AB$

Answer 1

The correct option is A $BE$

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 $A$ be $(0,h).$

Then, equation of $AC$ is $\frac{x}{a}+\frac{y}{h}=1$ ...(1)

and equation of $DE\perp AC$ and passing through origin is $\frac{x}{h}-\frac{y}{a}=0\Rightarrow x=\frac{hy}{a}$ ...(2)

On solving, Eqs. (1) and (2), we get the coordinates of point $E$ as follows $\frac{hy}{a^2}+\frac{y}{h}=1\Rightarrow y=\frac{a^{2}h}{a^2+h^2}$

$\therefore$ Coordinates of $E=(\frac{{ah}^2}{a^2+h^2},\frac{a^{2}h}{a^2+h^2})$

Since $F$ is mid point of $DE.$

$\therefore$ Coordinates of $F=(\frac{{ah}^2}{2(a^2+h^2)},\frac{a^{2}h}{2(a^2+h^2)})$

$\therefore$ Slope of $AF,$

$m_1=\frac{h-\frac{a^{2}h}{2(a^2+h^2)}}{0-\frac{{ah}^2}{2(a^2+h^2)}}=\frac{2h(a^2+h^2)-a^{2}h}{{-ah}^2}$

$\Rightarrow m_1=\frac{-(a^2+2h^2)}{ah}$ ...(3)

And slope of $BE,m_2=\frac{\frac{a^{2}h}{a^2+h^2}-0}{\frac{{ah}^2}{a^2+h^2}+a}=\frac{a^{2}h}{{ah}^2+a^3+{ah}^2}$

$\Rightarrow m_2=\frac{ah}{a^2+2h^2}$ ...(4)

From Eqs. (3) and (4),

$m_1{m}_2=-1\Rightarrow AF\perp BE$