Question

In a triangle $ABC,\angle B={90}^0,BE=3$ and $BD=4$. If $AE=ED=DC$, then the value of $AC$ can be expressed as $a\sqrt{b}$, where $a,b$ are positive integers and $b$ is square free. Find $a+b$.

Answer 1

Let $CD=DE=EA=x$

Applying Apollonius theorem in $△ABD$

${BD}^2+{AB}^2=2\{{BE}^2+x^2\}\Rightarrow 16+{AB}^2=2(9+x^2)\Rightarrow {AB}^2=2+2x^2$ .....(i)

Applying Apollonius theorem in $△BCE$, we have

${BE}^2+{BC}^2=2\{{BD}^2+x^2\}\Rightarrow 9+{BC}^2=2(16+x^2)\Rightarrow {BC}^2=23+2x^2$ .....(ii)

In right angled $△ABC$

${AB}^2+{BC}^2={AC}^2{AB}^2+{BC}^2={\left(3x\right)}^2$ .....(iii)

Substituting values of equations (i) and (ii) in equation (iii), we get

$2+2x^2+23+2x^2=9x^2\Rightarrow 25=5x^2\Rightarrow x=\sqrt{5}$$

Thus $AC=3x=3\sqrt{5}$

We have $a\sqrt{b}=3\sqrt{5}$

$\Rightarrow a=3,b=5\Rightarrow a+b=8$

Hence, the answer is $8$.