3. In a non-isosceles triangle $ABC$, the bisector $AD$ intersects the circumcircle of the triangle at point $P$. Point $I$ is the incenter of triangle $ABC$. It turns out that $ID = DP$. Find the ratio $AI: ID$. (20 points)
Solution: From the condition, it follows that $A I$ and $B I$ are the angle bisectors of triangle $A B C$. Let $\angle B A P = \angle C A P = \alpha$ and $\angle A B I = \angle C B I = \beta$. Then $\angle C B P = \alpha$, as an inscribed angle, and therefore $\angle I B P = \alpha + \beta$. Note that $\angle B I P = \angle B A P + \angle A B I = \alpha + \beta$, as the exterior angle of triangle $A I B$. This means that triangle $B I P$ is isosceles and $P I = P B$. Let $P D = D I = x$, then $P B = 2 x$. Describe a circle $\omega$ around triangle $A D B$. Since $\angle C B P = \angle P A B$, line $P B$ is tangent to circle

![](https://cdn.mathpix.com/cropped/2024_05_06_3ab34ed9f91a5fa18d1cg-2.jpg?height=959&width=1054&top_left_y=223&top_left_x=821)
$\omega$, so $P D \cdot P A = P B^2$. From this, $P A = 4 x$, $A I = P A - P I = 2 x$ and $A I : I D = 2 : 1$.

Answer: 2:1

Remark: The fact proven in the problem, that $P I = P B$ (and, accordingly, $P I = P C$), is known as the trident lemma. Referring to the lemma is accepted without proof.