G1.2 In Figure $1, A B C D$ is a parallelogram with $B A=3 \mathrm{~cm}$, $B C=4 \mathrm{~cm}$ and $B D=\sqrt{37} \mathrm{~cm}$. If $A C=h \mathrm{~cm}$, find the value of $h$.
$C D=B A=3 \mathrm{~cm}$ (opp. sides, $/ /$-gram)
In $\triangle B C D, \cos C=\frac{4^{2}+3^{2}-\sqrt{37}^{2}}{2 \times 3 \times 4}=-\frac{1}{2}$ $\cos B=\cos \left(180^{\circ}-C\right)=-\cos C=\frac{1}{2}$ (int. $\angle \mathrm{s} A B / / D C$ )
In $\triangle A B C, A C=\sqrt{3^{2}+4^{2}-2 \cdot 3 \cdot 4 \cos B}=\sqrt{13} ; h=\sqrt{13}$
Method 2 http://www2.hkedcity.net/citizen_files/aa/gi/fh7878/public_html/Trigonometry/Apollonius_Theorem.pdf By Apollonius theorem,
$$
\begin{array}{l}
2 B A^{2}+2 B C^{2}=A C^{2}+B D^{2} \\
2 \times 3^{2}+2 \times 4^{2}=h^{2}+37 \\
h=\sqrt{13}
\end{array}
$$