4. In triangle $A B C$, a circle $\omega$ with radius $r$ is inscribed, touching side $A B$ at point $X$. On the circle, point $Y$ is marked, diametrically opposite to point $X$. Line $C Y$ intersects side $A B$ at point $Z$. Find the area of triangle $A B C$, given that $C A + A Z = 1$.
Answer: $r$.

![](https://cdn.mathpix.com/cropped/2024_05_06_bd69057b1c245517edf6g-09.jpg?height=494&width=674&top_left_y=1575&top_left_x=768)

Solution. Draw a tangent to $\omega$ through point $Y$, intersecting segments $A C$ and $B C$ at points $P$ and $Q$ respectively. Let $M$ and $N$ be the points where $\omega$ touches sides $A C$ and $B C$ respectively. Notice that

$$
C P+P Y=C P+P M=C M=C N=C Q+Q N=C Q+Q Y
$$

Lines $A B$ and $P Q$ are parallel as perpendiculars to the same diameter. Therefore, triangles $C Q Y$ and $C B Z$, as well as $C P Y$ and $C A Z$, are similar with a coefficient $k=\frac{C Y}{C Z}$. Then

$$
C B+B Z=k(C Q+Q Y)=k(C P+P Y)=C A+A Z=1
$$

from which

$$
S_{A B C}=\frac{1}{2}(C A+A B+B C) \cdot r=\frac{1}{2}(C A+A Z+B Z+C B) \cdot r=r
$$