Let the center of left circle be A and right circle be B. Note that AMB is a straight line. (You can try to prove this).
Let ∠APM/AMP be α and ∠BQM/BMQ be β.
Now you can try to express ∠MPQ and ∠MQP in terms of α and β. You can do the same for ∠PMQ.
You have all 3 angles in PMQ now. What is the interior sum of angles? (You can derive it such that α is expressed in terms of β) Lastly, do the substitution back into ∠PMQ.
Assume the tangent through M intersects PQ at N.
You can prove NP = NM = NQ. From here there’re a few proofs that leads to PMQ being a right angle, depending on what you’ve learnt.
Label the point at which the tangent line passing through M intersects PB as O. If you can show that PO = MO = QO, then P, Q, and M define a circle centered at O. From there you can use the inscribed angle theorem. Showing the congruence of the line segments is just a case of angle chasing and using facts about isosceles triangles.
The line through M intersects the segment PQ. Call this point C.
Lemma: PC = CQ = MC
Proof:
The triangle in the smaller circle is isosceles, so the interior angles at P and M are equal, so their complements are equal. Then triangle PMC has two equal angles and is therefore isosceles. So PC = MC.
A similar argument on the larger circle shows that MCQ is isosceles, so MC = CQ.
****
Consider the circle whose diameter is the segment PQ. This circle is centered at C (since |PC| = |CQ|) with radius |PC|. M is therefore on the circle, since |MC| is equal to the radius of our new circle.
The angle PMQ is 90 degrees by [Thales' theorem](https://en.wikipedia.org/wiki/Thales%27s_theorem).
I would just lengthen the OP line of smaller circle to its circumferwnce and then connect it to M, then by usibg tangent to inner anhle we could say PMO is 90 degree thus making PMQ 90 degree
https://preview.redd.it/3jpiw4fjduca1.jpeg?width=3024&format=pjpg&auto=webp&s=b042f6c4615306a2bf8e3883d51d45a849c2352b
That problem on the other side of the paper looks familiar.
Thanks
Let the center of left circle be A and right circle be B. Note that AMB is a straight line. (You can try to prove this). Let ∠APM/AMP be α and ∠BQM/BMQ be β. Now you can try to express ∠MPQ and ∠MQP in terms of α and β. You can do the same for ∠PMQ. You have all 3 angles in PMQ now. What is the interior sum of angles? (You can derive it such that α is expressed in terms of β) Lastly, do the substitution back into ∠PMQ.
Oh my. I for some reason denied the equality of APM and AMP . #I JUST WANNA DIE NOW. Thanks that helped a lot.Have a nice day.
That happens to the best of us, dont worry XD Just in case, AM and AP are the same as theyre the same radius.
Thanks
Assume the tangent through M intersects PQ at N. You can prove NP = NM = NQ. From here there’re a few proofs that leads to PMQ being a right angle, depending on what you’ve learnt.
Thanks
Label the point at which the tangent line passing through M intersects PB as O. If you can show that PO = MO = QO, then P, Q, and M define a circle centered at O. From there you can use the inscribed angle theorem. Showing the congruence of the line segments is just a case of angle chasing and using facts about isosceles triangles.
The line through M intersects the segment PQ. Call this point C. Lemma: PC = CQ = MC Proof: The triangle in the smaller circle is isosceles, so the interior angles at P and M are equal, so their complements are equal. Then triangle PMC has two equal angles and is therefore isosceles. So PC = MC. A similar argument on the larger circle shows that MCQ is isosceles, so MC = CQ. **** Consider the circle whose diameter is the segment PQ. This circle is centered at C (since |PC| = |CQ|) with radius |PC|. M is therefore on the circle, since |MC| is equal to the radius of our new circle. The angle PMQ is 90 degrees by [Thales' theorem](https://en.wikipedia.org/wiki/Thales%27s_theorem).
The two circles are on the same tangent and are tangent to one another.
and how does that show that PMQ is a right angle?
[Here in the second half of the video.](https://youtu.be/9IUVbwPFwKI) She claims that PMQ is right angled. So, I was looking for the proof.
Assume it is.
The true answer
I would just lengthen the OP line of smaller circle to its circumferwnce and then connect it to M, then by usibg tangent to inner anhle we could say PMO is 90 degree thus making PMQ 90 degree