b) Let's construct point $E$ on side $AC = AB$, symmetric to $D$ with respect to the height dropped from $A$ - the axis of symmetry of triangle $ABC$. The sum of the angles we need to find is obviously half the sum of $\angle D K_{1} E + \angle D K_{2} E + \ldots + \angle D K_{n-1} E \quad(*)$

of the angles under which the segment $DE$ is seen from points $K_{1}, \ldots, K_{n-1}$, which divide the base $AC$ into $n$ equal parts.

Indeed, the sums of the angles under which the segments $AD$ and $AE$ are seen from these points are equal, and $\angle D K_{i} A + \angle A K_{i} E = \angle D K_{i} E$ for each $i=1,2, \ldots, n-1$. Since $B D E K_{1}, K_{1} D E K_{2}, K_{2} D E K_{3}, \ldots, K_{n-1} D E C$ are parallelograms, the sum $(*)$ can easily be "assembled" into one angle $B D K_{n-1}$, equal to $\angle A = \alpha$ (see figure):

$\angle B D K_{1} = \angle D K_{1} E, \angle K_{1} D K_{2} = \angle D K_{2} E, \ldots, \angle K_{n-2} D K_{n-1} = \angle D K_{n-1} E$.

![](https://cdn.mathpix.com/cropped/2024_05_06_f7e07b07084714df1adcg-36.jpg?height=474&width=512&top_left_y=1881&top_left_x=773)

## Answer

$\alpha / 2$.

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Proivolov V.V.

In an isosceles triangle $ABC (AB = AC)$, the angle $A$ is equal to $\alpha$. On the side $AB$, a point $D$ is taken such that $AD = AB / n$. Find the sum of $n-1$ angles under which the segment $AD$ is seen from the points dividing the side $BC$ into $n$ equal parts:

a) for $n=3$

b) for an arbitrary $n$.