## Task 32/73

To construct a triangle from the heights $h_{a}$ and $h_{b}$ and the angle bisector $w_{\gamma}$ (using the usual notation.)
Assuming the triangle is already constructed. Then it holds (Figure)

$$
\frac{A D}{D B}=\frac{A G}{B H}=\frac{A C}{B C}=\frac{b}{a}=\frac{h_{a}}{h_{b}}
$$

due to $p_{1}\left\|p_{2}\right\| p_{3}$, the intercept theorem, the theorem on the ratio of the segments created by the angle bisector, and because $a \cdot h_{a}=b \cdot h_{b}=2 A$.

From this follows the construction:

1. Draw a pair of parallel lines $p_{1}$ and $p_{2}$ at a distance $h: a$!
2. Draw a parallel line $p_{3}$ between $p_{1}$ and $p_{2}$ such that the distance from $p_{1}$ to $p_{3}$ is in the ratio $h_{a}: h_{b}$ to the distance from $p_{3}$ to $p_{2}$.
3. Choose a point $C$ arbitrarily on $p_{1}$ and draw a circle with radius $C D=w_{\gamma}$ around $C$. The intersection points with $p_{3}$ are $D$ and $D^{\prime}$.
4. At $C$, measure the acute angle formed by $C D$ and $p_{1}$ and transfer this angle to $C D$ such that the free leg does not coincide with $p_{1}$. The intersection of the free leg with $p_{2}$ is $A$.
5. Draw the line through $A$ and $D$; its intersection with $p_{1}$ is $B$.

Determination: The point $D^{\prime}$ yields a congruent triangle with the opposite orientation. The construction is always feasible if $w_{\gamma}>h_{a} \cdot h_{b}:\left(h_{a}+h_{b}\right)$; only then does a point $D$ with $\measuredangle D C B<90^{\circ}$ arise. Except for the mentioned congruence, the construction is then also always unique.