Problem 5.

Quadrilateral $ABCD$ is inscribed in a circle with a diameter of 17. Diagonals $AC$ and $BD$ are perpendicular. Find the sides $AB, BC, CD$, given that $AD=8$ and $AB:CD=3:4$.
Solution. Triangles $A O B$ and $C O D$ are similar by two angles ($A O B, C O D$ - vertical, $A B D, A C D$ - subtend the same arc), hence $A O: O D=A B: C D=3: 4$.

Let $A O=3 y, \quad O D=4 y$, then, by the Pythagorean theorem for the right triangle $A O D$ we have $9 y^{2}+16 y^{2}=A D^{2}=64, \quad$ from which $\quad y=1.6, \quad A O=4.8$, $O D=6.4, \sin \angle O D A=\frac{A O}{A D}=0.6$.

Let $A B=3 x, C D=4 x$. By the Law of Sines for

![](https://cdn.mathpix.com/cropped/2024_05_06_67444cb6f337a86f9dddg-2.jpg?height=637&width=697&top_left_y=744&top_left_x=1268)
triangle $A B D$ we have $\frac{A B}{\sin \angle B D A}=2 R$, that is $\frac{3 x}{0.6}=17, x=3.4, A B=10.2, C D=13.6$. By the Pythagorean theorem for the right triangles $A O B, C O D$ and $B O C$ we find $B O=9, C O=12, B C=15$.

Answer: $A B=10.2, C D=13.6, B C=15$.