In the isosceles triangle $A B C$, $A C B \angle = A B C \angle = 40^{\circ}$. On the ray $A C$, we measure and mark a segment $A D$ equal to $B C$. What is the measure of $\angle B D C$?
Let $E$ be the point in the plane such that the triangle $ABC$ is congruent to the triangle $EDA$, and $E$ is on the same side of the line $AD$ as $B$. This point $E$ clearly exists uniquely. Then:

$$
AE = ED = AB = AC
$$

and

$$
\angle ADE = \angle EAD = \angle ABC = \angle ACB = 40^\circ.
$$

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Since the sum of the angles in a triangle is $180^\circ$, we have $\angle BAC = 100^\circ$, so

$$
\angle BAE = \angle BAC - \angle EAD = 100^\circ - 40^\circ = 60^\circ.
$$

Thus, the isosceles triangle $EAB$ is also equilateral, i.e.,

$$
EA = AB = AE = ED.
$$

Therefore, point $E$ is the center of the circumcircle of triangle $ABD$. $\angle BDC = \angle BDA$ is a circumferential angle subtended by the arc $AB$, and thus equal to half the central angle of $AB$, which is $30^\circ$.

Remark. We remind our readers of the note in Exercise 2267 (Vol. 36, p. 20, January 1986). If the intersection of $OP_3$ and $P_1P_6$ is $B$, and the intersection of $OP_8$ and $P_5P_{10}$ is $C$, then the point system $O, B, C, P_3$ is congruent to the system $A, B, C, D$ here, and thus $\angle BDC = \angle P_{12}P_3P_9 = \angle P_{12}OP_9 / 2 = 30^\circ$. Prove the facts used!