## Task 11/64

For a quadrilateral $A B C D$, the following segments are given: $A B=a, \measuredangle B C A=\gamma_{1}, \measuredangle A C D=\gamma_{2}$, $\measuredangle C D B=\delta_{1}, \measuredangle B D A=\delta_{2}$.

a) The quadrilateral is to be constructed.

b) How large is the side $C D=c$, if $A B=a=1$ (LE), $\gamma_{1}=30^{\circ}, \gamma_{2}=30^{\circ}$, $\delta_{1}=45^{\circ}$ and $\delta_{2}=60^{\circ}$? (Solution by calculation)
a) Analysis: The triangle $A C D$ is determined up to similarity by the angles $A C D=\gamma_{2}$ and $C D A=\delta=\delta_{1}+\delta_{2}$. The triangle $B C D$ is determined up to similarity by the angles $B D C=\delta_{2}$ and $B C D=\gamma=\gamma_{1}+\gamma_{2}$. Therefore, the quadrilateral $A B C D$ is also determined up to similarity.

To construct a quadrilateral $A^{\prime} B^{\prime} C^{\prime} D^{\prime}$ similar to $A B C D$, one first constructs a side $D^{\prime} C^{\prime}$ of arbitrary length and marks the angles $C D B$ and $C D A$ at $D^{\prime}$, and the angles $A C D$ and $B C D$ at $C^{\prime}$. The intersection points of the free legs are $A^{\prime}$ and $B^{\prime}$, respectively.

Then, a similarity transformation is performed such that the side $A^{\prime} B^{\prime}$ becomes the side $A B=a$.

Construction Description: One fixes the segment $C^{\prime} D^{\prime}$ arbitrarily and marks the angle $A C D=\gamma_{2}$ at $C^{\prime}$ and the angle $C D A=\delta=\delta_{1}+\delta_{2}$ at $D^{\prime}$. The intersection of the free legs is $A^{\prime}$. Then, one marks the angle $B C D=\gamma=\gamma_{1}+\gamma_{2}$ at $C^{\prime}$ and the angle $C D B=\delta_{1}$ at $D^{\prime}$, and the intersection of the free legs is $B^{\prime}$.

The similarity transformation can be performed by marking the segment $A B=a$ on $A^{\prime} B^{\prime}$ from $A^{\prime}$ and drawing a parallel to $B^{\prime} C^{\prime}$ through $B$; the intersection of this parallel with the line through $A^{\prime}$ and $C^{\prime}$ is $C$. One then draws a parallel through $C$ to $C^{\prime} D^{\prime}$, which intersects the line through $A^{\prime} D^{\prime}=A D^{\prime}$ at $D$.

Determination: All sub-constructions are always and uniquely possible if the sum of any three of the four given angles is not equal to $180^{\circ}$; otherwise, there are no intersection points of the free legs, and the problem has no solution.

b) Based on the solution in a), it is expedient to perform the calculation on a quadrilateral $A^{\prime} B^{\prime} C^{\prime} D^{\prime}$ similar to $A B C D$. Then it holds that

$$
\frac{C D}{A B}=\frac{C^{\prime} D^{\prime}}{A^{\prime} B^{\prime}} \quad ; \quad C D=\frac{C^{\prime} D^{\prime}}{A^{\prime} B^{\prime}} A B
$$

Since $C^{\prime} D^{\prime}$ is arbitrary, it is expedient to choose it to be 1 (LE). Then, because $A B=1$ (LE): $C D=\frac{1}{A^{\prime} B^{\prime}}$. Using the sine rule of plane trigonometry, one first calculates the segments $D^{\prime} A^{\prime}$ and $D^{\prime} B^{\prime}$ from $C^{\prime} D^{\prime}=1$ and the given angles, and then the segment $A^{\prime} B^{\prime}$ using the cosine rule. It is

$$
\begin{gathered}
\measuredangle D A C=\alpha_{1}=180^{\circ}-\gamma_{2}-\delta_{1}-\delta_{2}=45^{\circ} \text { and } \\
\measuredangle D B C=\beta_{2}=180^{\circ}-\gamma_{1}-\gamma_{2}-\delta_{1}=75^{\circ} \\
D^{\prime} A^{\prime}=D^{\prime} C^{\prime} \cdot \frac{\sin \gamma_{2}}{\sin \alpha_{1}}=\frac{\sin \gamma_{2}}{\sin \alpha_{1}}=\frac{\sqrt{2}}{2} \approx 0.7071 \\
D^{\prime} B^{\prime}=D^{\prime} C^{\prime} \cdot \frac{\sin \gamma_{1}+\gamma_{2}}{\sin \beta_{1}}=\frac{\sin \gamma_{1}+\gamma_{2}}{\sin \beta_{2}}=\frac{3 \sqrt{2} \sqrt{6}}{2} \approx 0.8966
\end{gathered}
$$

$\sin \beta_{2}=\sin 75^{\circ}$ was calculated here elementarily using the addition theorem.

$$
A^{\prime} B^{\prime}=\sqrt{\left(D^{\prime} A^{\prime}\right)^{2}+\left(D^{\prime} B^{\prime}\right)^{2}-2 \cdot D^{\prime} A^{\prime} \cdot D^{\prime} B^{\prime} \cdot \cos \delta_{2}}=\ldots=\frac{1}{2} \sqrt{20-10 \sqrt{3}}
$$

From $C D=\frac{1}{A^{\prime} B^{\prime}}$ it follows finally that $C D=\frac{2}{\sqrt{20-10 \sqrt{3}}} \approx 1.22$.