Problem 815

Answer: $38^{\circ}$.

Solution. Let $a$ be the length of the lateral side of the trapezoid. Note that point $T$ lies on the perpendicular bisector of the bases of the trapezoid, that is, on its axis of symmetry. From symmetry, we get that $B T=T C=a$ (Fig. 5).

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Fig. 5: to the solution of problem 10.4

Next, note that $\angle B A D=\angle C D A=2 \angle C A D$, so $A C$ is the bisector of angle $B A D$. Since $\angle C A D=\angle A C B$ due to parallelism, triangle $A B C$ is isosceles. Therefore, $B C$ is also equal to $a$.

We have obtained that triangle $B T C$ is equilateral, and its angles are each $60^{\circ}$. Now it is not difficult to calculate the required angle:

$$\angle T C D=\angle B C D-\angle B C T=\left(180^{\circ}-\angle A D C\right)-60^{\circ}=120^{\circ}-\angle A D C=38^{\circ} .$$