We present another excersise of Euclidean geometry. It is a “classical” problem on the “$$80-80-20$$” isosceles triangle.

Given the isosceles triangle $$\triangle ABC$$ with $$\angle ABC = \angle CAB = 80°$$, fix a point $$P$$ on the side $$AC$$ such that $$CP\cong AB$$. What is the measure of the angle $$\angle CPB$$?

In order to take advantage of the congruence between $$AB$$ and $$CP$$ we need to draw some other triangles. For example, we can construct two replicas of $$\triangle ABC$$ adjacent to each other, as in the Figure below, where $$\triangle ABC\cong \triangle BCD \cong \triangle CDE$$.

1. Triangle $$\triangle ACE$$ is isosceles by construction. What can you say about the angle $$\angle ACE$$, though?
2. Draw then your conclusions on triangle $$\triangle ACE$$.
3. Determine, by subtraction, the measure of $$\angle EAB$$.
4. The information you have is enough to state that $$\triangle CPB$$ and $$\triangle ABE$$ are congruent. Thanks to what congruence criterion?
5. Your aim is now to calculate the measure of $$\angle CPB\cong\angle ABE$$.
6. Concentrate on $$\triangle BDE$$. What kind of triangle is that? Why? Use this observation to compute the measure of $$\angle DBE$$.
7. By correclty using the measure of known angles, determine that of $$\angle ABE$$, and, from that, your answer. You should find $$\angle CPB = 150°$$.

As an exercise, try to demonstrate the same assertion again, using, this time, the construction below, where $$\triangle ADB$$ is equilateral.