Awesome geometry -6

First, note that $DAE=90^{\circ}$ because the degree measure between the internal and external angle bisectors is always $90^{\circ}$ .

Now, draw triangle $ADE$ . It's a right triangle at vertex $A$ , AND $AD=60,AE=80$ . Thus by Pythagorean Theorem $DE=100$ .

With $A,D,E$ drawn, let's draw $B$ and $C$ . They are just points on the line $DE$ such that $\angle CAD = \angle BAD$ .

We are given the area of $\triangle ABC$ , so if we knew the height from $A$ to $BC$ then we would know $BC$ .

However, this height is the same height from $A$ to $DE$ , and that can be easily computed by finding the area of triangle $ADE$ in two different ways: $[ADE]=\dfrac{80\cdot 60}{2} = \dfrac{100\cdot h}{2}\implies h =48$

Now we have $\dfrac{h\cdot BC}{2}=[ABC]\implies\dfrac{48\cdot BC}{2}=696\implies BC=\boxed{29}$ and we're done.

$DAE = 90^o, \text {since it is the angle between internal and external bisectors. }\\$ $\\ \therefore~DAE \text{ is a right angled triangle}\\ \text{with sides 60 and 80, and hypotenuses }DE. \\ \therefore DE = \sqrt{60^2 + 80^2} =100. ~\therefore~Sin(ADB)= \dfrac{80}{100}\\$ $\text{ area of the triangle } ABC = \dfrac{1}{2}*BC*60*Sin(ADB)\\ = \dfrac{1}{2}*BC*60*\dfrac{80}{100}\\=696~$ $\implies~BC= 696*\dfrac{2}{1}*\dfrac{100}{80*60} = \boxed{ 29 }$