Two Tangents Revisited

From right triangle $ODA$ by Pythagorean theorem $AD=\sqrt{13^2-5^2}=12$ . Also $\angle OAD=arcsin( \frac{5}{13})=22.62^\circ$ .

Designating $CD=CE=x$ we get $BE=BF=7-x$ .

So $AC=AD-CD=12-x$ and $AB=AF-BF=12-(7-x)=5+x$ .

Law of cosines for $\triangle ABC$ will then be $BC^2=AC^2+AB^2-2\times AC\times AB\times cos(2\times 22.62^\circ)$

$49=(12-x)^2+(5+x)^2-2\times (12-x)\times (5+x)\times cos(45.24^\circ)$

There are two solutions. (Together they add up to 7, of course.) The larger will give us the smaller of the lengths $AC$ and $AB$ , and it is $x=4.854$ . $AC=12-x=7.146$ .