11.4. Quadrilateral $ABCD$ is inscribed in a circle, $BC=CD, AC=c, \angle BAD=2\alpha$. Find the area of this quadrilateral.
Answer: $c^{2} \sin \alpha \cos \alpha$.

Solution. Since the chords $B C$ and $C D$ are equal, the arcs $B C$ and $C D$, not containing point $A$, are also equal. Therefore, $\angle B A C=\angle D A C=\alpha$.

First method. Let $\angle A B C=\beta, \angle A D C=\gamma$. Translate triangle $A B C$ so that the equal segments $B C$ and $C D$ coincide, with point $A$ moving to point $E$ (see Fig. 11.4a). Since quadrilateral $A B C D$ is inscribed, $\beta+\gamma=180^{\circ}$, which means points $A, D$, and $E$ lie on the same straight line.

Then the area of quadrilateral $A B C D$ is equal to the area of triangle $A C E$. In triangle $A C E: A C=C E$ and $\angle A=\angle E=\alpha$, so $\angle A C E=180^{\circ}-2 \alpha$.

Therefore, $S_{A B C D}=S_{A C E}=\frac{1}{2} A C^{2} \sin \left(180^{\circ}-2 \alpha\right)=\frac{1}{2} c^{2} \sin 2 \alpha=$

![](https://cdn.mathpix.com/cropped/2024_05_06_51c28cf2652dc1e7d646g-3.jpg?height=705&width=403&top_left_y=227&top_left_x=1523)
$c^{2} \sin \alpha \cos \alpha$.

Second method. From triangles $A B C$ and $A D C$ using the cosine theorem: $B C^{2}=A B^{2}+A C^{2}-2 A B \cdot A C \cos \alpha$ and $C D^{2}=A D^{2}+A C^{2}-2 A D \cdot A C \cos \alpha$. Since $B C=C D$, we have $A B^{2}+A C^{2}-2 A B \cdot A C \cos \alpha=A D^{2}+A C^{2}-2 A D \cdot A C \cos \alpha \Leftrightarrow A B^{2}-A D^{2}=2 A C \cdot(A B-A D) \cos \alpha$.

If $A B \neq A D$, then from the obtained equality, it follows that $A B+A D=2 A C \cos \alpha$. Then $S_{A B C D}=S_{\triangle A B C}+S_{\triangle A D C}=\frac{1}{2} A B \cdot A C \sin \alpha+\frac{1}{2} A D \cdot A C \sin \alpha=$ $\frac{1}{2} A C(A B+A D) \sin \alpha=c^{2} \sin \alpha \cos \alpha$.

If $A B=A D$, then triangles $A B C$ and $A D C$ are equal. The sum of angles $A B C$ and $A D C$ is $180^{\circ}$, which means each of these angles is a right angle ($A C$ is the diameter of the circle). Then $A B=A D=c \cdot \cos \alpha$, so $S_{A B C D}=2 \cdot \frac{1}{2} c \cos \alpha \cdot c \sin \alpha=c^{2} \sin \alpha \cos \alpha$.

![](https://cdn.mathpix.com/cropped/2024_05_06_51c28cf2652dc1e7d646g-3.jpg?height=388&width=354&top_left_y=1294&top_left_x=1571)

The equality $A B+A D=2 A C \cos \alpha$ can also be obtained using Archimedes' formula. Let the bisector of the inscribed angle $BAD$ intersect the circle at point $C$, and let $K$ be the projection of point $C$ onto the line $A B$. Then $A K=\frac{A D+A B}{2}$ (see Fig. 11.4b). Now, from the right triangle $AKC$, we get that $A K=A C \cos \alpha$.

Third method. Let point $E$ lie on the circle and $D E \| A C$ (see Fig. 11.4c). Then $A C D E$ is an inscribed isosceles trapezoid.

Note that $S_{\triangle A C D}=S_{\triangle A C E}$ (these triangles have equal bases and heights). Then $S_{\text {ABCD }}=S_{\triangle A B C}+S_{\triangle C D A}=S_{\triangle A B C}+S_{\triangle A E C}=S_{\text {ABCE }}$. Since $A E=C D=B C$, $A E C B$ is an inscribed isosceles trapezoid.

Then $B E=A C$, and the angle $B F C$ between them is equal to half the sum of the smaller arcs $B C$ and $A E$, which is $2 \alpha$.

Therefore, $S_{A B C D}=S_{A B C E}=\frac{1}{2} A C^{2} \sin 2 \alpha=c^{2} \sin \alpha \cos \alpha$.

![](https://cdn.mathpix.com/cropped/2024_05_06_51c28cf2652dc1e7d646g-3.jpg?height=449&width=377&top_left_y=1997&top_left_x=1545)

Grading criteria.

“+” A complete and well-reasoned solution (the formula of Archimedes can be used without proof) “士” A generally correct reasoning, containing minor gaps or inaccuracies (for example, in the second method, the case $A B=A D$ is not considered) “干” Correct ideas are present, but the solution is not fully developed

“-” Only a special case is considered

“-” Only the answer is provided

“-” The problem is not solved or is solved incorrectly