29. As shown in the figure, in $\triangle \mathrm{ABC}$, $\mathrm{AB}=13$ cm, $\mathrm{AC}=10$ cm, and $\mathrm{AH} \perp \mathrm{BC}$. Take point D between $\mathrm{H}$ and $\mathrm{C}$, such that $\angle B A D=60^{\circ}, \angle D A C=2 \angle H A D$. Then $\mathrm{BH}: H C=$ $\qquad$ -
【Analysis】As shown in the figure, fold $\triangle \mathrm{BHA}$ along $\mathrm{AB}$ to get $\triangle \mathrm{ABE}$, then fold $\triangle \mathrm{ABE}$ along $\mathrm{AE}$ to get $\triangle \mathrm{AEF}$.  
Since $\angle A E F=\angle A E B=\angle A H B=90^{\circ}$, points $\mathbf{B} 、 \mathbf{E} 、 \mathbf{F}$ are collinear.  
And $\angle F A H=3 \angle B A H, \angle C A H=3 \angle D A H, \angle B A H+\angle D A H=60^{\circ}$,  
so $\angle F A H+\angle C A H=180^{\circ}$.  
Therefore, points A, C are collinear.  
Thus, CFB is a triangle.

Let $S_{\triangle A B H}=13 a$  
Then $S_{\triangle A B F}=26 a$  
$$
\begin{array}{l}
\because A F: A C=A B: A C=13: 10 \\
\therefore S_{\triangle A B C}=20 a \\
\therefore S_{\triangle A H C}=7 a \\
\therefore B H: H C=S_{\triangle A H B}: S_{\triangle A H C}=13: 7
\end{array}
$$