6. (8 points) On the circumscribed circle of isosceles triangle $A B C(A C=$ $B C$ ), a point $P$ is taken (point $P$ lies on the arc $A B$ not containing point $C$). Point $D$ is the foot of the perpendicular dropped from point $C$ to the line $B P$. Find the length of the segment $P D$, if $A P=a$ and $B P=b$.

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# Solution:

![](https://cdn.mathpix.com/cropped/2024_05_06_f48d5baeb4f84f12ed51g-06.jpg?height=883&width=857&top_left_y=398&top_left_x=611)

In all cases, the parameters $a$ and $b$ were given such that $\angle C A C^{\prime}=90^{\circ}$, and therefore the height $C D^{\prime}$ falls on the extension of $A P$ beyond point $A$.

Since quadrilateral $A B C D$ is cyclic, $\angle C B P=\angle C A D^{\prime}$. It is also known that $A C=B C$, hence $\triangle C B D=\triangle C A D^{\prime}$ (by hypotenuse and acute angle). Therefore, $B D=A D^{\prime}$ and $C D=C D^{\prime}$. Triangles $\triangle C P D$ and $\triangle C P D^{\prime}$ are also equal by the leg $\left(C D=C D^{\prime}\right)$ and the common hypotenuse $C P$. Thus, $P D^{\prime}=P D$. In the end, we get

$$
\begin{gathered}
P D=P B-D B=P B-D^{\prime} A=P B-\left(P D^{\prime}-P A\right)= \\
=b+a-P D^{\prime}=b+a-P D
\end{gathered}
$$

From which it follows that $P D=\frac{a+b}{2}$.

## Answers:

| Variant | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| Answer | 4.9 | 4.9 | 3.95 | 4.8 | 5.15 | 4.65 | 4.55 | 4.95 | 3.9 | 5.35 |