Find the length of a line from the a vertex to a midpoint of a triangle

Apply the cosine rule to the triangles ABD and ACD

\begin{align} &4^2=4^2+AD^2-8AD\cos\angle ADB\\ &6^2=4^2+AD^2-8AD\cos(180-\angle ADB) \end{align}

Note that $\cos(180-x)=-\cos x$ and add the two equations to get $AD^2=10$.

We have $$CD^2=\frac{c^2}{4}+b^2-2b\frac{c}{2}\cos(\alpha)$$ and $$a^2=b^2+c^2-2bc\cos(\alpha)$$ you can eliminate $\alpha$ to get $CD^2$

Through the Pythagorean theorem you may prove that in a parallelogram the sum of the squared lengths of the sides equals the sum of the squared lengths of the diagonals. It follows that in a triangle with side lengths $a,b,c$ the length $m_A$ of the median through $A$ fulfills $$ m_A^2 = \frac{2b^2+2c^2-a^2}{4}.$$

You can use the cosine rule and it would be really simple, but if you don't want to use it or don't know about it then you can use the Heron's formula, Pythagoras theorem and the rule of Median (that is, a median divides the opposite side of a triangle into two equal parts).

Let's start, Let there be a triangle ABC with D is the midpoint of BC. Join A to D. So here AD is the median of triangle ABC on BC. A median divides the opposite side into two equal parts. So, BD=DC=4. Now use the heron's formula to calculate the area of ABC.

$S=\frac{(AB+BC+CA)}{2}$ $S=\frac{4+6+8}{2}=9$

Heron's formula states that,

$Ar.(ABC)=\sqrt{S(S-AB)(S-BC)(S-CA)}$

Putting the values, $Ar.(ABC)=\sqrt{9(9-4)(9-6)(9-8)}=\sqrt{9×5×3×1}=3\sqrt{15}$

Now, area of ABC is also equals to

$Ar.(ABC)=\frac{1}{2}×base×height=\frac{1}{2}×8×height$

$3\sqrt{15}=4×height$

$height=\frac{3}{4}\sqrt{15}$

Let AE be the height of the triangle on BC. By using Pythagoras theorem on triangle ABE

$AB^2=BE^2+AE^2$

$BE^2=(4)^2-(\frac{3}{4}\sqrt{15})^2$

$BE=\sqrt{\frac{256-135}{16}}=\sqrt{\frac{121}{16}}=\frac{11}{4}$

$BD-BE=ED$

$ED=4-\frac{11}{4}=\frac{16-11}{4}=\frac{5}{4}$

Now,

$AD^2=AE^2+ED^2$

$AD=\sqrt{\frac{135+25}{16}}=\sqrt{frac{160}{16}}=\sqrt{10}$

I hope this is helpful.