Finding the Ratio of Segments of a Triangle In $\triangle ABC$, $AB$ = $5$, $BC = 7$, $AC = 9$, and $BD= 5$. Determine the ratio $AD:DC$ if I have to represent my answer as the ratio $k:w$ where $k$ and $w$ are relatively prime positive integers. 
I know the answer I must come to is $19:8$.
 
 A: Use Stewart's Theorem:
$$\frac{AB^2}{AD.AC}-\frac{BD^2}{AD.DC}+\frac{BC^2}{CD.AC}=1$$
$$\frac{5^2}{9AD}-\frac{5^2}{AD.DC}+\frac{7^2}{9CD}=1\quad (1)$$
Now call $$\frac{AD}{DC}=k→\frac{AD+DC}{DC}=k+1→\frac{9}{DC}=k+1→DC=\frac{9}{k+1}\quad (2)$$
And
$$AD= \frac{9k}{k+1}\quad (3)$$
Now replace $(2)$ and $(3)$ in $(1)$ and finish the calculations.
A: The ratios of the areas of the three triangles are equal to the ratios of the lengths of the bottom edges. If we work with a known area, we can find $AD$, and then $AD:DC=AD:(9-AD)$.  
We can use Heron's formula to compute the areas. For the outer triangle, $S_1=(5+7+9)/2=\frac{21}2$ and $A_1=\sqrt{S_1(S_1-AB)(S_1-BC)(S_1-AC)}=\frac{21}4\sqrt{11}$. Setting $x$ to be the length $AD$, we have for the inner isosceles triangle $S_2=(x+5+5)/2$ and $A_2=\sqrt{S_2(S_2-x)(S_2-5)^2}$. Set $A_2/A_1=AD/AC=x/9$ and square both sides to eliminate the radical. This gives you a quartic equation for $x$, but an $x^2$ can be factored out leaving an easily-solved quadratic equation. Use the constraint that $0\lt x\lt9$ to eliminate the extraneous solution.
A: By the cosine law we have 
$$\cos A =\frac{25+81-49}{2\cdot 5\cdot 9}$$
and this can help you find $AD,$ which I don't think should be $5\sqrt2.$
