## Task 5 - 020625

Draw a line segment $A B=5 \mathrm{~cm}$! Construct a $45^{\circ}$ angle $\alpha$ at point $A$ on line segment $A B$!

On the ray that does not contain point $B$, find a point $P$ with the following property:

If you connect $P$ and $B$, then $\angle A B P=\angle A P B$.

How can this point $P$ be constructed?
The points $A, B$ and $P$ form a triangle where the two angles at $B$ and $P$ are equal. Therefore, the triangle $\triangle A B P$ is isosceles, with the base being $B P$ and the legs being $A B = A P$. Thus, the point $P$ can be found by marking off the length of the segment $\overline{A B} = 5$ $\mathrm{cm}$ on the second leg.