Problem 9.5. Let $A B C D$ be a cyclic quadrilateral with circumcircle $k$. The rays $\overrightarrow{D A}$ and $\overrightarrow{C B}$ meet at point $N$ and the line $N T$ is tangent to $k, T \in k$. The diagonals $A C$ and $B D$ meet at the centroid $P$ of $\triangle N T D$. Find the ratio $N T: A P$.

Ivailo Kortezov
9.5. It follows from the condition that $T$ lies on the $\operatorname{arc} B C$. Let $M=N T \cap D P$ be the midpoint of $N T$. Then we have

$$
M B \cdot M D=M T^{2}=M N^{2}
$$

Thus $M B: M N=M N: M D$ and it follows that $\triangle N M B \sim \triangle D M N$. Hence

$$
\Varangle M N B=\Varangle M D N=\Varangle N C A
$$

![](https://cdn.mathpix.com/cropped/2024_06_03_f8ccd71b4db367e7c2a9g-147.jpg?height=521&width=552&top_left_y=1392&top_left_x=1137)

i.e. $N T \| A C$. Thus

$$
\frac{N T}{A P}=\frac{2 N M}{A P}=\frac{2 M D}{P D}=3
$$