2. Let $\triangle A B C$ be an obtuse isosceles triangle with base $\overline{A B}$. On the base, there is a point $D$ such that $|\angle A C D|=35^{\circ}$. On the leg $\overline{B C}$, there is a point $E$ such that $|C E|=|C D|$. Determine $|\angle B D E|$.
2. Given is an obtuse isosceles triangle $\triangle A B C$ with base $\overline{A B}$. On the base, there is a point $D$ such that $|\angle A C D|=35^{\circ}$. On the leg $\overline{B C}$, there is a point $E$ such that $|C E|=|C D|$. Determine $|\angle B D E|$.

## Solution.

In an obtuse isosceles triangle, the obtuse angle is the angle opposite the base.

Therefore, $|\angle A C B|>35^{\circ}$.

Sketch:

![](https://cdn.mathpix.com/cropped/2024_05_30_ba00de53bdad84e80dffg-16.jpg?height=717&width=1465&top_left_y=1035&top_left_x=250)

In the given isosceles triangle $\triangle A B C$, the side $\overline{A B}$ is the base, so the angles at the base are congruent:

$|\angle B A C|=|\angle C B A|=\alpha$

Triangle $\triangle C D E$ is also isosceles because $|C E|=|C D|$.

Therefore, $|\angle E D C|=|\angle C E D|=\beta$.

The measure of an exterior angle of a triangle is equal to the sum of the measures of the opposite interior angles of the triangle.

Thus:

$\beta=|\angle B D E|+\alpha$ and

$|\angle B D E|+\beta=\alpha+35^{\circ}$.

By substituting $\beta$ from the first equation into the second equation, we get:

$|\angle B D E|+|\angle B D E|+\alpha=\alpha+35^{\circ}$

$2|\angle B D E|=35^{\circ}$

$|\angle B D E|=35^{\circ}: 2=17.5^{\circ}=17^{\circ} 30^{\prime}$.