Prove that $\Delta BPQ$ is an isosceles triangle

Question

Given was the following figure:

Also the following were given:

• $M_1$ and $M_2$ are the centres of the two circles
• The two circles have the same radius

First, I added an other line through the points $B$ and $Q$. This line will intersect with the $M_1$-circle and called this point $N$. After that, I have proven the following:

$$\angle PQB = \angle AQN \ \ (\text{vertical angle})$$

$$\angle APB = \angle ANB \ \ (\text{inscribed angle})$$

$$\Delta AQN \sim \Delta BQP \ \ (\text{AAA})$$

So, I figured out that if I prove that $\angle BPQ = \angle BQP$, then the triangle has to be an isosceles triangle. The only problem is, I don't know how to continue or if I'm on the right path.

Do you have any hints for proving that $\angle BPQ = \angle BQP$?

Answer 1

Hint: Look at the angle $\angle BAP$ as inner angle of the two circles. Then use the fact that the two circles have the same radius.

Answer 2

For the ones wondering, I did the following:

$$2 \cdot \angle BAP = \angle BM_1P$$

$$2 \cdot \angle BAP = \angle BM_2Q$$

The length of $BP = 2r_1 \cdot \sin(\frac12\angle BM_1P) = 2r_1 \cdot \sin(\angle BAP)$

The length of $BQ = 2r_2 \cdot \sin(\frac12\angle BM_2Q) = 2r_2 \cdot \sin(\angle BAP)$

Conclusion, since $r_1 = r_2$, $BP$ has to be the same value as $BQ$ and $\Delta BPQ$ has to be an isosceles triangle.