23. Let $A B C$ be a triangle such that the altitude from $A$, the median from $B$, and the internal angle bisector from $C$ meet at a single point. If $B C=10$ and $C A=15$, find $A B^{2}$.
23. 205
Let $D$ be the foot of the $A$-altitude, $E$ the midpoint of $A C$, and $F$ the foot of the $C$-internal angle bisector. Then by Ceva's Theorem, we have $\frac{A F}{F B} \cdot \frac{B D}{D C} \cdot \frac{C E}{E A}=1$, and so $\frac{b}{a} \cdot \frac{c \cos B}{b \cos C}=1$, where we are using the shorthand $B C=a, C A=b, A B=c$. By cosine law, we know that $\cos C=\frac{a^{2}+b^{2}-c^{2}}{2 a b}$ and $\cos B=\frac{c^{2}+a^{2}-b^{2}}{2 c a}$. Substituting this into the equation, we obtain
$$
a\left(a^{2}+b^{2}-c^{2}\right)=b\left(c^{2}+a^{2}-b^{2}\right)
$$

Solving for $c^{2}=A B^{2}$ in this equation then gives the final answer, which is 205 .