3.12. Two heights of a parallelogram, drawn from the vertex of the obtuse angle, are equal to $h_{1}$ and $h_{2}$, and the angle between them is $\alpha$. Find the larger diagonal of the parallelogram.
3.12. According to the condition, in parallelogram $ABCD$ we have: $BF \perp AD, BE \perp CD, BF = h_{1}, BE = h_{2}, \angle FBE = \alpha$ (Fig. 3.14). Then $\angle ABE = 90^{\circ}$, and $\angle BAF = \angle FBE$ as the acute angles with mutually perpendicular sides. Draw CKLAD;

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-506.jpg?height=340&width=633&top_left_y=1172&top_left_x=522)

Fig. 3.14 we have $AC^{2} = (AD + DK)^{2} + CK^{2}$ (from $\triangle AKC$), $AD = BC = \frac{h_{2}}{\sin \alpha}$ (from $\triangle BEC$), $DK = CK \operatorname{ctg} \alpha = \frac{h_{1} \cos \alpha}{\sin \alpha}$ (from $\triangle CKD$). Therefore,

$$
AC^{2} = \left(\frac{h_{2}}{\sin \alpha} + \frac{h_{1} \cos \alpha}{\sin \alpha}\right)^{2} + h_{1}^{2} =
$$

$$
\begin{gathered}
= \frac{h_{2}^{2} + 2 h_{1} h_{2} \cos \alpha + h_{1}^{2} \cos^{2} \alpha + h_{1}^{2} \sin^{2} \alpha}{\sin^{2} \alpha} = \\
= \frac{h_{2}^{2} + 2 h_{1} h_{2} \cos \alpha + h_{1}^{2}}{\sin^{2} \alpha}, \text{ hence } AC = \frac{\sqrt{h_{1}^{2} + 2 h_{1} h_{2} \cos \alpha + h_{2}^{2}}}{\sin \alpha}
\end{gathered}
$$

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-507.jpg?height=330&width=669&top_left_y=458&top_left_x=54)

Fig. 3.15