$QP$ isn't the radius of the circle, but how would you show that it has the same length?

Question

Exercise In the figure, $P$ is a point inside the circle of radius 1 and distinct from the center. $\triangle PAB$ is an equilateral triangle. Besides, $BA=BC$. If the segment $CP$ is extended to cut the circumference at $Q$, show that $PQ=1$.

Another geometry problem, but this time it has a circle. I was trying to draw some segments from the ratio to other points of the figure. and well i know that $\Delta PBC$ is an isosceles triangle.I don't have more ideas, i'd appreciate any help. It's not homework, it's just a problem i'm curious to. Thanks!

Answer 1

Since $AB=BC$, we have $$ m\angle AQB=m\angle BQC=m\angle ACB =m\angle BAC. $$ Let $\theta$ be the common measure of these four angles. Now argue:

1. Triangle $ABC$ is isosceles, so $m\angle ABC=180-2\theta$.

2. $m\angle ABP=60$, so $m\angle PBC=120 -2\theta$.

3. Triangle $PBC$ is isosceles, so $m\angle BPC=\frac12(180-m\angle PBC)=30+\theta$.

4. $m\angle QPA+m\angle APB+m\angle BPC=180$, so $m\angle QPA=90-\theta$.

5. $m\angle AQP=m\angle AQC=2\theta$, so the remaining angle in triangle $AQP$ has measure $m\angle QAP=90-\theta$.

6. So triangle $AQP$ is isosceles with apex angle $2\theta$.

7. Let $O$ be the center of the circle. Triangle $AOB$ is isosceles. By the inscribed angle theorem, its apex angle has measure $m\angle AOB=2\theta$.

8. Sides $AP$ and $AB$ are congruent, so the triangles $AQP$ and $AOB$ are congruent. It follows that $QP=OB$.