Ptolemy's Riddle V

With $AC$ as a diameter, we can conclude that the angles $\angle ABC$ and $\angle ADC$ are right angles, due to Thales' theorem . With this piece of information, we can construct $3$ more congruent copies of this quadrilateral $ABCD$ and put them all together as a perfect square. That is, their opposite angles are complementary (adding up to $180 ^\circ$ ), forming straight line lengths of the square and they comprise the right angles as the square's vertices while the isosceles sides edge with one another as the intersecting perpendicular segments at the centroid, as shown below:

Thus far, it is clear that this cyclic quadrilateral is, in fact, a quarter of a square, whose side length is the sum of $AB + BC$ and whose diagonal is twice the length of $BD = 6$ : The big square has a diagonal length of $12$ .

Thus, the area of the big square = $\dfrac{1}{2}(12^2) = 72$ .

Finally, the area of $ABCD$ = $\dfrac{1}{4}(72) = \boxed{18}$ .

By Thales' Theorem, the fact that the lower triangle is isosceles, that AC is the $x$ axis, AC is 7 in length and the circle's center is at the origin without loss of generality establishes the coordinates of the lower triangle's vertices. A circle with its center at D with a radius of 6 (given in the problem) intersects the original circle at $\left\{\pm\frac{1}{7} \left(6 \sqrt{13}\right),\frac{23}{14}\right\}$ . To be consistent with the diagram, I used the point with the negative $x$ coordinate value. See Area - Coordinate Geometry in Irregular Polygons . The area is 18.

By Thales' Theorem , $\angle ABC$ and $\angle ADC$ are right angles, and since $AC = 7$ and $AD = CD$ , by Pythagorean's Theorem on $\triangle ACD$ we obtain $AD = CD = \frac{7\sqrt{2}}{2}$ , so the area of $\triangle ACD$ is $A_{\triangle ACD} = \frac{49}{4}$ , and by Pythagorean's Theorem on $\triangle ABC$ we obtain the relation $AB^2 + BC^2 = 49$ .

By Ptolemy's Theorem , $AC \cdot BD = AB \cdot CD + AD \cdot BC$ . Substituting $AC = 7$ , $BD = 6$ , and $AD = CD = \frac{7\sqrt{2}}{2}$ we obtain $7 \cdot 6 = AB \cdot \frac{7\sqrt{2}}{2} + \frac{7\sqrt{2}}{2} \cdot BC$ which simplifies to $AB + BC = 6\sqrt{2}$ .

Combining $AB^2 + BC^2 = 49$ and $(AB + BC)^2 = (6\sqrt{2})^2$ gives $AB \cdot BC = \frac{23}{2}$ , so the area of $\triangle ABC$ is $A_{\triangle ABC} = \frac{1}{2} \cdot AB \cdot BC = \frac{1}{2} \cdot \frac{23}{2} = \frac{23}{4}$ .

The area of quadrilateral $ABCD$ is the sum of the areas of the triangles $\triangle ABC$ and $\triangle ACD$ , so $A_{ABCD} = A_{\triangle ABC} + A_{\triangle ACD} = \frac{23}{4} + \frac{49}{4} = \boxed{18}$ .