4. In an isosceles triangle $A B C$ the angle opposite the base $\overline{A B}$ is $80^{\circ}$. On the height to the base of this triangle, a point $D$ is chosen such that $|\angle D B A|=30^{\circ}$. On the segment $\overline{B D}$, a point $T$ is chosen such that $|\angle T A D|=20^{\circ}$. Determine the size of the angle $\angle T C A$.
4. In the isosceles triangle $A B C$, the angle opposite the base $\overline{A B}$ is $80^{\circ}$. A point $D$ is chosen on the height to the base of this triangle such that $|\angle D B A|=30^{\circ}$. A point $T$ is chosen on the segment $\overline{B D}$ such that $|\angle T A D|=20^{\circ}$. Determine the size of the angle $\angle T C A$.

## Solution.

$\triangle A B C$ is isosceles, and $\angle A C B$ is the angle opposite the base, so $|\angle A C B|=80^{\circ}$.

Since $|\angle A C B|=80^{\circ}$ is the angle opposite the base of the isosceles triangle $A B C$, it follows that $|A C|=|B C|$ and the sizes of the other two angles of this triangle are $|\angle B A C|=|\angle C B A|=\left(180^{\circ}-80^{\circ}\right): 2=50^{\circ}$.

Let point $N$ be the foot of the perpendicular from $C$ to the base $A B$ of the triangle.

![](https://cdn.mathpix.com/cropped/2024_05_30_ad9a4c9088848eb03059g-11.jpg?height=460&width=700&top_left_y=2049&top_left_x=1135)
$A B C$.

Since the height to the base of an isosceles triangle bisects the base, the line $N C$ is the perpendicular bisector of the base of triangle $A B C$, so $|A D|=|B D|$, which means that triangle $A B D$ is also isosceles. Therefore, $|\angle B A D|=|\angle D B A|=30^{\circ}$.

This means that $|\angle B D A|=180^{\circ}-2 \cdot 30^{\circ}=120^{\circ}$.

Consider triangle $A T D$.

$|\angle D T A|=180^{\circ}-\left(120^{\circ}+20^{\circ}\right)=40^{\circ}$.

Determine the size of the angle $\angle B A T$.

$|\angle B A T|=|\angle B A D|-|\angle T A D|$

![](https://cdn.mathpix.com/cropped/2024_05_30_ad9a4c9088848eb03059g-12.jpg?height=448&width=699&top_left_y=424&top_left_x=1158)

$|\angle B A T|=30^{\circ}-20^{\circ}=10^{\circ}$.

Now determine the angles of triangle $A D C$.

Since $C N$ is the perpendicular bisector of the base of the isosceles triangle $A B C$, it is also the angle bisector of the angle opposite the base, so $|\angle A C D|=80^{\circ}: 2=40^{\circ}$.

$|\angle D A C|=|\angle B A C|-|\angle B A D|$

$|\angle D A C|=50^{\circ}-30^{\circ}=20^{\circ}$.

Thus, $|\angle C D A|=180^{\circ}-\left(40^{\circ}+20^{\circ}\right)=120^{\circ}$.

It follows that $\triangle A T D \cong \triangle A D C$

![](https://cdn.mathpix.com/cropped/2024_05_30_ad9a4c9088848eb03059g-12.jpg?height=476&width=719&top_left_y=1107&top_left_x=951)

because they share the side $\overline{A D}$ and have corresponding congruent angles adjacent to it (SAS congruence criterion).

From this congruence, it follows that $|A T|=|A C|$, so triangle $A T C$ is isosceles.
![](https://cdn.mathpix.com/cropped/2024_05_30_ad9a4c9088848eb03059g-12.jpg?height=868&width=1566&top_left_y=1551&top_left_x=247)

The angle opposite the base of this triangle is $\angle T A C$ and it holds that $|\angle T A C|=50-10^{\circ}=40^{\circ}$, so the size of the desired angle $|\angle T C A|=\left(180^{\circ}-40^{\circ}\right): 2=70^{\circ}$.