## Task $7 / 61$

Given is an isosceles trapezoid with the bases $a$ and $b$ and the height $h$.

a) Construct the point $P$ on the axis of symmetry from which the two legs appear at a right angle.

b) Determine the distance of the point $P$ from one of the two bases algebraically.

c) Under what conditions is the construction of the point $P$ possible? (Discussion of possible cases)
a) Analysis: $E P_{1}=x_{2}, F P_{1}=x_{1}, A B=a, D C=b, E F=h$, and $\measuredangle B P_{1} C=\measuredangle B P_{2} C=1 R$. It follows: $P_{1}, P_{2}$ must lie on the Thales circle over $B C$ or $A D$.

Construction and Determination:

The Thales circle over one of the two legs is drawn. This circle intersects the axis of symmetry either in two points $\left(P_{1}, P_{2}\right)$, one point $(P)$, or no point. These points are the ones sought after.

b) It holds (see figure): $\triangle B F P_{1} \sim \triangle C E P_{1}$. From this it follows

$$
x_{1}: \frac{a}{2}=\frac{b}{2}: x_{2} \quad \text { i.e. } \quad x_{1} \cdot x_{2}=\frac{a b}{4}
$$

Furthermore, it holds: $x_{1}+x_{2}=h$ (2). From (1) and (2) it follows that $x_{1}$ and $x_{2}$ are solutions of the equation $x^{2}-h x+\frac{a b}{4}=$ 0

$$
x_{1 ; 2}=\frac{h}{2} \pm \frac{1}{2} \sqrt{h^{2}-a b}
$$

c) According to the considerations in b: If $h^{2}>a b$, there are two solutions, i.e., two points; if $h^{2}=a b$, there is one solution, i.e., one point, and if $h^{2}<a b$, there is no solution, i.e., no point. The construction is thus only possible if $h^{2} \geq a b$.