Let $t$ be TNYWR.

In the diagram, $C$ lies on $A E$ and $A B=B C=C D$. If $\angle C D E=t^{\circ}, \angle D E C=(2 t)^{\circ}$, and $\angle B C A=\angle B C D=x^{\circ}$, determine the measure of $\angle A B C$.

![](https://cdn.mathpix.com/cropped/2024_04_17_385ef00c66fdf6396726g-2.jpg?height=355&width=830&top_left_y=1476&top_left_x=669)
Consider $\triangle C D E$ and its exterior angle, $\angle A C D$.

We have $\angle A C D=\angle C D E+\angle C E D$ or $(2 x)^{\circ}=t^{\circ}+(2 t)^{\circ}$ which gives $2 x=3 t$.

Now $\triangle B A C$ is isosceles with $A B=B C$, so $\angle B A C=\angle B C A=x^{\circ}$.

Thus, $\angle A B C=180^{\circ}-\angle B A C-\angle B C A=180^{\circ}-(2 x)^{\circ}$.

Since $2 x=3 t$, then $\angle A B C=180^{\circ}-(3 t)^{\circ}$.

Since the answer to (b) is 20 , then $t=20$, and so $\angle B A C=180^{\circ}-60^{\circ}=120^{\circ}$.