Using the Pythagorean Theorem: In combination with the triangle area

The area of triangle ABC is equal to:

$\frac{AD\times BC}{2}=2x+16$

As we are given the area of the triangle, we can insert this data into BC in the formula:

$\frac{AD\times(BD+DC)}{2}=2x+16$

$\frac{AD\times(x+5+3)}{2}=2x+16$

$\frac{AD\times(x+8)}{2}=2x+16$

We then multiply by 2 to eliminate the denominator:

$AD\times(x+8)=4x+32$

Divide by: $(x+8)$

$AD=\frac{4x+32}{(x+8)}$

We rewrite the numerator of the fraction:

$AD=\frac{4(x+8)}{(x+8)}$

We simplify to X + 8 and obtain the following:

$AD=4$

We now focus on triangle ADC and by use of the Pythagorean theorem we should find X:

$AD^2+DC^2=AC^2$

Inserting the existing data:

$4^2+(x+5)^2=(\sqrt{65})^2$

$16+(x+5)^2\text{ }=65/-16$

$(x+5)^2=49/\sqrt{}$

$x+5=\sqrt{49}$

$x+5=7$

$x=7-5=2$

The area of triangle ABC is:

$\frac{AB\times AC}{2}=S$

Into this formula, we insert the given data:

$\frac{AB\times(x+4)}{2}=4x+16$

$\frac{AB\times(x+4)}{2}=4(x+4)$

Notice that X plus 4 on both sides is reduced, and we are left with the equation:

$\frac{AB}{2}=4$

We then multiply by 2 and obtain the following:

$AB=4\times2=8$

If we now observe the triangle ABC we are able to find side BC using the Pythagorean Theorem:

$AB^2+AC^2=BC^2$

We first insert the existing data into the formula:

$8^2+(x+4)^2=BC^2$

We extract the root:

$BC=\sqrt{64+x^2+2\times4\times x+4^2}=\sqrt{x^2+8x+64+8}=\sqrt{x^2+8x+72}$

We can now calculate AD by using the formula to calculate the area of triangle ABC:

$S_{\text{ABC}}=\frac{AD\times BC}{2}$

We then insert the data:

$4x+16=\frac{AD\times\sqrt{x^2+8x+80}}{2}$

$AD=\frac{(4x+16)\times2}{\sqrt{x^2+8x+80}}=\frac{8x+32}{\sqrt{x^2+8x+80}}$