## Task B-2.6.

Rectangle $A B C D$, in which $|A B|>|A D|$, rotates around vertex $D$ to form rectangle $E F G D$ such that vertex $A$ maps to point $E$ on the diagonal $\overline{B D}$. The intersection of sides $\overline{D C}$ and $\overline{E F}$ is point $H$. The area of rectangle $A B C D$ is in the ratio 5 : 2 to the area of quadrilateral $B C H E$. If $\varphi=\varangle A D B$, calculate $\frac{\sin ^{2} \varphi+1}{\cos ^{2} \varphi+1}$.
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_0c173970003c85556d5eg-11.jpg?height=734&width=574&top_left_y=889&top_left_x=684)

Triangle $D E H$ is similar to triangle $A B D$ by the AA similarity theorem.

From $P_{A B C D}: P_{B C H E}=5: 2$ it follows that $P_{A B D}: P_{B C H E}=5: 4$, 1 point, or $\frac{P_{A B D}}{P_{D E H}}=5: 1=k^{2}$.

Thus, the similarity coefficient is $k=\sqrt{5}$.

$\frac{|A B|}{|D E|}=\frac{\sqrt{5}}{1}$,

$\operatorname{tg} \varphi=\frac{|A B|}{|A D|}=\sqrt{5}$.

$$
\begin{aligned}
\frac{\sin ^{2} \varphi+1}{\cos ^{2} \varphi+1} & =\frac{\sin ^{2} \varphi+\cos ^{2} \varphi+\sin ^{2} \varphi}{\cos ^{2} \varphi+\cos ^{2} \varphi+\sin ^{2} \varphi} \\
& =\frac{2 \sin ^{2} \varphi+\cos ^{2} \varphi}{2 \cos ^{2} \varphi+\sin ^{2} \varphi}=\frac{2 \operatorname{tg}^{2} \varphi+1}{2+\operatorname{tg}^{2} \varphi}=\frac{2 \cdot 5+1}{2+5}=\frac{11}{7}
\end{aligned}
$$