Problem 10.4. An isosceles trapezoid $ABCD$ with bases $BC$ and $AD$ is such that $\angle ADC = 2 \angle CAD = 82^{\circ}$. Inside the trapezoid, a point $T$ is chosen such that $CT = CD, AT = TD$. Find $\angle TCD$. Give your answer in degrees.

![](https://cdn.mathpix.com/cropped/2024_05_06_85a336fa11c4e8eb26a1g-10.jpg?height=261&width=506&top_left_y=606&top_left_x=468)
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).

![](https://cdn.mathpix.com/cropped/2024_05_06_85a336fa11c4e8eb26a1g-10.jpg?height=394&width=515&top_left_y=1109&top_left_x=469)

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:

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