Question

In the following figure, $AE\perp BC$, D is the mid point of BC, then $x$ is equal to

• A. $\frac{1}{a}[b^2-d^2-\frac{a^2}{4}]$
• B. $\frac{h+d}{3}$
• C. $\frac{c+d-h}{2}$
• D. $\frac{a^2+b^2+d^2+c^2}{4}$

Answer 1

The correct option is A $\frac{1}{a}[b^2-d^2-\frac{a^2}{4}]$

In $△ABC$, $AD$ is the median.

By Apollonius theorem,

$A{B}^2+A{C}^2=2A{D}^2+2D{C}^2$

$c^2+b^2=2d^2+2{\left(\frac{1}{2}BC\right)}^2$

$c^2+b^2=2d^2+\frac{a^2}{2}$

But, $c^2=h^2+{(\frac{a}{2}-x)}^2$

$\therefore h^2+{(\frac{a}{2}-x)}^2+b^2=2d^2+\frac{a^2}{2}$

$h^2+\frac{a^2}{4}-ax+x^2+b^2=2d^2+\frac{a^2}{2}$

But $h^2=d^2-x^2$

$d^2-x^2-ax+x^2+b^2=2d^2+\frac{a^2}{4}$

$ax=b^2-d^2-\frac{a^2}{4}$

$x=\frac{1}{a}[b^2-d^2-\frac{a^2}{4}]$

Hence proved.