## Task A-3.3.

In triangle $A B C$, $|A B|=|A C|$. On side $\overline{A C}$, there is a point $D$ such that $|A D|<|C D|$, and on segment $\overline{B D}$, there is a point $P$ such that $\varangle A P C$ is a right angle. If $\varangle A B P=\varangle B C P$, determine $|A D|:|C D|$.
## First Solution.

Let $\varangle A B P = \varangle B C P = \varphi$ and $\varangle P B C = \varangle P C A = \psi$.

Let point $T$ be the midpoint of segment $\overline{A C}$. Since $\overline{A C}$ is the hypotenuse of the right triangle $A P C$, it follows that $|A T| = |P T| = |C T|$.

Therefore, $\varangle C P T = \varangle P C A = \psi$.

![](https://cdn.mathpix.com/cropped/2024_05_30_7166a15529367baa5955g-14.jpg?height=613&width=897&top_left_y=850&top_left_x=568)

Since $\varangle D P C$ is an exterior angle of triangle $B C P$, we have

$$
\varangle D P C = \varangle P C B + \varangle P B C = \varphi + \psi
$$

so $\varangle D P T = \varphi$.

By applying the Law of Sines in triangles $A B D$ and $P D T$, we get

$$
\frac{|A D|}{|A B|} = \frac{\sin \varangle A B D}{\sin \varangle A D B} \quad \text{and} \quad \frac{|D T|}{|P T|} = \frac{\sin \varangle D P T}{\sin \varangle P D T}
$$

Thus,

$$
\frac{|A D|}{|A B|} = \frac{\sin \varangle A B D}{\sin \varangle A D B} = \frac{\sin \varphi}{\sin (180^\circ - \varangle A D B)} = \frac{\sin \varangle D P T}{\sin \varangle P D T} = \frac{|D T|}{|P T|}
$$

From this, it follows that

$$
\frac{|A D|}{|A C|} = \frac{|A D|}{|A B|} = \frac{|D T|}{|P T|} = \frac{|A T| - |A D|}{|P T|} = \frac{\frac{1}{2}|A C| - |A D|}{\frac{1}{2}|A C|} = 1 - 2 \cdot \frac{|A D|}{|A C|}
$$

Thus, $|A D| : |A C| = 1 : 3$.

Finally, the desired ratio is $|A D| : |C D| = 1 : 2$.