4. In the square $A B C D$ with side length $6 \mathrm{~cm}$, a point $T$ is given on the diagonal $\overline{A C}$ such that

(8) $|T C|=\frac{1}{4}|A C|$. The line through points $B$ and $T$ intersects the side $\overline{C D}$ at point $P$.

Determine the distance $|P C|$.
Solution. Triangles $A B T$ and $C P T$ are similar (equal angles) and it holds that

$$
|A B|:|P C|=|A T|:|T C|=\frac{3}{4}|A C|: \frac{1}{4}|A C|=3: 1
$$

thus $|P C|=\frac{1}{3}|A B|=2$. The distance $|P C|$ is $2 \mathrm{~cm}$.

![](https://cdn.mathpix.com/cropped/2024_05_30_fac652c8a845a57c6fa7g-02.jpg?height=531&width=466&top_left_y=563&top_left_x=795)

(full solution 8 points)

(ratios set 4 points)