A geometry problem by Akshat Sharda

$\text{Let ABC be the triangle, AB=AC=8, BC=4.}\\ AD~ altitude~ of~ ~ISOSCELES~~\Delta.~~~~~~BE ~the~ angle~ bisector. \\ Cos\alpha=\dfrac 2 8 =\dfrac 1 4.\\ \therefore by~ tri.~ formulii, Sin\alpha=\dfrac{\sqrt{15}} 4,~~\\ Sin\dfrac \alpha 2=\sqrt{\dfrac 3 8},~~~Cos\dfrac \alpha 2=\sqrt{\dfrac 5 8}\\ \therefore~Sin(\alpha+\dfrac \alpha 2) =Sin\alpha*Cos\dfrac \alpha 2 + Cos\alpha *Sin\dfrac \alpha 2=\dfrac{3\sqrt3}{4\sqrt2}...(**)\\\text{Applying Sin Law in }\Delta~ EBC,\\ \dfrac{BE}{Sin\alpha} =\dfrac 4 {SinCEB}=\dfrac 4 {(**)}\\ \therefore~~~BE= \dfrac{ \sqrt{15}} 4 *\dfrac 4 {\frac{3\sqrt3}{4\sqrt2} }\\ = \dfrac 4 3 *\sqrt{10}=\dfrac a c *\sqrt{b} \\ \therefore~~a+b+c=~~~~~~\large \color{#D61F06}{17}$

For solving this,

We have a great formula , $M_{a}=\huge \frac{2\sqrt{b×c×s(s-a)}}{b+c}$

Where :

$M_{a}$ = Median to side $a$ .

$s$ = Semi-perimeter.

Let , $a=b=8$ and $c=4$

Therefore ,

$\Rightarrow M_{a}=\frac{2\sqrt{8×4×10×2}}{8+4}$

$\Rightarrow M_{a}=\frac{16\sqrt{10}}{12}=\boxed{\frac{4\sqrt{10}}{3}}$