Let's construct an isosceles triangle if the base and the length of the angle bisector of the base angle are given as \(a\) and \(f\). Specifically, consider the cases \(a = f\) and \(a = f \cdot \sqrt{2}\).

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Construct the isosceles triangle if the base and the length of the angle bisector of the base angle are given as \(a\) and \(f\). Specifically, consider the cases \(a = f\) and \(a = f \cdot \sqrt{2}\).
We denote the endpoints of the base by $B$ and $C$, the opposite vertex by $A$, the point where the angle bisector from $C$ intersects $AB$ by $F$, and measure on the extension of $BC$ beyond $B$ the segment

$$
BG = BF = g
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_4fc9d8e3b211abba4af3g-1.jpg?height=658&width=805&top_left_y=392&top_left_x=649)

The resulting triangle $BFG$ is isosceles, and the angles at the base are equal to half the angle at $B$ in triangle $ABC$. Therefore, $CFG$ is also isosceles and similar to $GFB$. Thus, $BG : GF = FC : CG$, or

$$
g : f = f : (g + a)
$$

This is a quadratic equation in $g$, whose roots are

$$
g_{12} = -\frac{a}{2} \pm \sqrt{\left(\frac{a}{2}\right)^2 + f^2}
$$

Thus, (1) always has exactly one positive root:

$$
g = \sqrt{\left(\frac{a}{2}\right)^2 + f^2} - \frac{a}{2}
$$

Measure the segment $BE = f$ perpendicular to $BC = a$, and draw a circle through $E$ with the midpoint $D$ of $BC$ as its center. Let this circle intersect $BC$ at points $G$ and $H$ (with $GB < \frac{a+g}{2} - g > 0$), which is equivalent to

$$
\frac{2}{3} a < f < \sqrt{2} a
$$

This is the condition for constructibility, and it is also the condition for the existence of a solution.

The above holds for the special cases $f = a$ and $f = a / \sqrt{2}$. In the first case, $BE = BC$, so $E$ and $C$ are points on the circle of radius $a$ in which a regular decagon is inscribed, and one of its sides is the triangle $BCF$. We can recognize the classical construction of the segment $BF = g$ in the above solution, which we have generalized to the case $f \neq a$.

If $f = a / \sqrt{2}$, then (2) implies

$$
a - g = (3 - \sqrt{3}) \frac{a}{2} = \sqrt{3} g
$$

Thus, $\angle DBF = 30^\circ$.

Remark. For the angle $\delta$ at the base of triangle $CGF$,

$$
\cos \delta = \frac{CG}{2GF} = \frac{a + g}{2f}
$$

Thus, we get a quadratic equation for $\cos \delta$ similar to the one for $g$. Most solvers chose this path.