Balti Tee 2025 · Ülesanne 12

Let $E=AI\cap(ABC)$, where $(ABC)$ denotes the circumcircle of $ABC$. Then $E$ lies on $AI$ and $EA=ET$.

We have $DQ\parallel ET$, since

$$\angle ADQ=\angle APQ=\angle APT=\angle AET.$$

Therefore triangles $IDQ$ and $IET$ are similar, giving

$$\frac{ID}{IE}=\frac{DQ}{ET}.$$

It is enough to show

$$\frac{ID}{IE}=\frac{AI}{AE}.$$

Subtracting $1$ from both sides gives the equivalent relation

$$\frac{ED}{IE}=\frac{EI}{AE},$$

or

$$EI^2=ED\cdot EA.$$

This is a standard power relation: $EI=EB$, and $EB$ is tangent to $(ABD)$. Hence the desired equality follows.