27. In the diagram below, $A B C$ is a triangle such that $A B$ is longer than $A C$. The point $N$ lies on $B C$ such that $A N$ bisects $\angle B A C$. The point $G$ is the centroid of $\triangle A B C$, and it is given that $G N$ is perpendicular to $B C$. Suppose $A C=6 \mathrm{~cm}, B C=5 \sqrt{3} \mathrm{~cm}$ and $A B=x \mathrm{~cm}$. Find the value of $x$.
27. Answer: 9
Let $B C=a, C A=b$ and $A B=c$. We shall prove that $c+b=a \sqrt{3}$. Thus $c=5 \sqrt{3} \times \sqrt{3}-6=9$. Using the angle bisector theorem, we have $B N / N C=c / a>1$ so that $B N>B M$. Also $\angle A N C=\angle B+\frac{1}{2} \angle A<\angle C+\frac{1}{2} \angle A<\angle A N B$ so that $\angle A N C$ is acute. This shows that $B, M, N, D$ are in this order, where $M$ is the midpoint of $B C$ and $D$ is the foot of the perpendicular from $A$ onto $B C$.

First we have $B D=c \cos B=\frac{a^{2}+c^{2}-b^{2}}{2 a}$. Using the angle bisector theorem, $B N=$ $\frac{a c}{b+c}$. Therefore, $M N=B N-\frac{a}{2}=\frac{a c}{b+c}-\frac{a}{2}=\frac{a(c-b)}{2(c+b)}$. Also $M D=B D-\frac{a}{2}=\frac{a^{2}+a^{2}-b^{2}}{2 a}-$ $\frac{a}{2}=\frac{c^{2}-b^{2}}{2 a}$. Since $G N$ is parallel to $A D$ and $G$ is the centroid of the triangle $A B C$, we have $M D / M N=3$. It follows that $c+b=a \sqrt{3}$. Thus, $A B=a \sqrt{3}-b=15-6=9$.