10.053. In parallelogram $ABCD$, the height drawn from vertex $B$ of the obtuse angle to side $DA$ divides it in the ratio 5:3, counting from vertex $D$ (Fig. 10.53). Find the ratio $AC:BD$, if $AD:AB=2$.
Solution.

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-581.jpg?height=444&width=292&top_left_y=94&top_left_x=110)

Fig. 10.54
![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-581.jpg?height=542&width=754&top_left_y=93&top_left_x=480)

Fig. 10.56

Given $\frac{A D}{A B}=2, \frac{E D}{A E}=\frac{5}{3} ; A D=3 x+5 x=8 x, \frac{8 x}{A B}=2$, $A B=4 x, B E=\sqrt{16 x^{2}-9 x^{2}}=x \sqrt{7}, B D=\sqrt{25 x^{2}+7 x^{2}}=x \sqrt{32}=4 x \sqrt{2}$ $\left(\triangle B E D, \angle B E D=90^{\circ}\right)$. In $\triangle A F C \quad \angle A F C=90^{\circ}, A C=\sqrt{F C^{2}+A F^{2}}$; $A F=B E=x \sqrt{7}, \quad F C=3 x+8 x=11 x, \quad A C=\sqrt{121 x^{2}+7 x^{2}}=8 x \sqrt{2} ;$ $\frac{A C}{B D}=\frac{8 x \sqrt{2}}{4 x \sqrt{2}}=2$.

Answer: $2: 1$.