Point on the Bisector

Apply Law of cosines. First equation, triangle ABD; $BD^2=5^2 + AD^2 -2*5*AD*\cos 30^\circ$ . Second eq., triangle ACD; $CD^2= 25^2 + AD^2 -2*25*AD*\cos 30^\circ$ . As $CD=BD$ , solve the equation by silmutaneous and we get $AD = \frac {25^2 - 5^2} { (25 - 5) \sqrt{3} } = 10 \sqrt{3}$ . This gives $AD^2= 300$ .

On drawing the triangle with some precision, we note D is external to BC. Drop normals BP & CQ on AD. Clearly, angles $BAD=CAD=30^\circ$ . So $BP=AB \sin 30^\circ = 5 \sin 30^\circ$ . Similarly, $AP=5 \cos 30^\circ$ , $AQ=25 \cos 30^\circ$ , $CQ=25 sin 30^\circ$ . Consider $BD=DC=x$ . Now, applying Pythagoras' Theorem, $AD= 5 \cos 30^\circ + \sqrt{x^2-(5 sin 30^\circ)^2} = 25 \cos 30^\circ + \sqrt{x^2-(25 \sin 30^\circ)^2}$ On solving, we get: $x=5 \sqrt{7}$ & subsequently $AD= 10 \sqrt{3}$ So $AD^2 = 300$ .

The point $D$ is uniquely determined as the intersection of the perpendicular bisector of $BC$ and the angle bisector of $\angle BAC$ . From the law of cosines, we have $BD^2 = 5 ^2 +AD^2 - 2 \times 5 \times AD \times \cos 30^\circ$ and $DC^2=25 ^2+AD^2-2\times 25 \times AD \times \cos 30^\circ$ .

Since $BD = DC$ , putting the two equations together we have $AD= \frac {25 ^2- 5 ^2} { (25 -5) \times \sqrt{3} } = 10 \sqrt{3}$ . Thus $AD^2 = 300$ .