## Task 3 - 160913

Let $\triangle A B C$ be an isosceles triangle with base $A B$, length $\overline{A B}=b$, and leg length 2b. The points $D$ and $E$ are the interior points of $A C$ and $B C$, respectively, where the legs intersect the circle with diameter $A B$. Determine the perimeter of the quadrilateral $A B E D$.
The midpoint of $AB$ is $M$. Then triangle $DAM$ is isosceles with $DM = AM = \frac{b}{2}$.

Therefore, $\angle MDA = \angle MAD = \angle BAC = \angle ABC$, so $\triangle DAM \sim \triangle ABC$.

Consequently, $DA : AM = AB : BC = 1 : 2$, thus $DA = \frac{1}{2} AM = \frac{b}{4}$, and therefore

$$
CD = AC - DA = 2b - \frac{b}{4} = \frac{7b}{4}
$$

Similarly, we obtain $EB = \frac{b}{4}$ and $CA = \frac{7b}{4}$.

Thus, $DE \parallel AB$ and consequently $\triangle DEC \sim \triangle ABC$. This gives $DE : AB = CD : AC = 8 : 7$, so $DE = \frac{7b}{8}$. Therefore, the desired perimeter is

$$
AB + BE + ED + DA = b + \frac{b}{4} + \frac{7b}{4} + \frac{b}{4} = \frac{19b}{8}
$$

![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-0987.jpg?height=571&width=334&top_left_y=2056&top_left_x=861)