MHB Prove that S lies on the line AB

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The discussion revolves around proving that point S lies on line AB in a geometric configuration involving a rhombus PQRS with specific angle measures. Participants analyze the tangential properties of lines QP and RM, and how they relate to angles formed by the circumcircle of triangle PSR. The key argument is based on the theorem that states the angle between a tangent and a chord equals the angle subtended by the chord, leading to the conclusion that angles ASP and RSB are both 60 degrees. Some participants express confusion about the implications of these angles, while others clarify that if M is the circumcenter of triangle PSR, the tangents can be shown to intersect correctly. Ultimately, the conversation highlights the geometric relationships and theorems that support the proof.
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Let $PQRS$ be a rhombus with $\angle Q=60^{\circ}$. $M$ is a point inside triangle $PSR$ such that $\angle PMR=120^{\circ}$. Let lines $QP$ and $RM$ intersect at $A$ and lines $QR$ and $PM$ intersect at $B$. Prove that $S$ lies on the line $AB$.
 
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anemone said:
Let $PQRS$ be a rhombus with $\angle Q=60^{\circ}$. $M$ is a point inside triangle $PSR$ such that $\angle PMR=120^{\circ}$. Let lines $QP$ and $RM$ intersect at $A$ and lines $QR$ and $PM$ intersect at $B$. Prove that $S$ lies on the line $AB$.

Since $PQRS$ is a rhombus, we have $\angle PQR=\angle PSR=60^{\circ}$.

It can be easily seen that $M$ is the circumcentre of $\Delta PSR$ and $\Delta PSR$ is an equilateral triangle.

Also, $QPA$ and $QRB$ are tangents to the circumcircle of $\Delta PSR$.

Hence, $\angle PSA=60^{\circ}$ and $\angle RSB=60^{\circ}$.

The angles $\angle PSA,\angle PSR$ and $\angle RSB$ sum to $180^{\circ}$, hence $S$ lies on$AB$.
 
Pranav said:
Since $PQRS$ is a rhombus, we have $\angle PQR=\angle PSR=60^{\circ}$.

It can be easily seen that $M$ is the circumcentre of $\Delta PSR$ and $\Delta PSR$ is an equilateral triangle.

Also, $QPA$ and $QRB$ are tangents to the circumcircle of $\Delta PSR$.

Hence, $\angle PSA=60^{\circ}$ and $\angle RSB=60^{\circ}$.

The angles $\angle PSA,\angle PSR$ and $\angle RSB$ sum to $180^{\circ}$, hence $S$ lies on$AB$.

Thanks for participating, Pranav!:)

But...I don't follow your reasoning because I don't understand why the equilateral triangle $PSR$ and that $QPA$ and $QRB$ are tangents to the circumcircle of $\Delta PSR$ imply $\angle PSA=60^{\circ}$ and $\angle RSB=60^{\circ}$.

I am definitely not saying your approach is wrong, it just that I don't get it, could you elaborate more on that, please? :o
 
Given the line ${AB}$ is tangent to the circle at $S$, there is a theorem that states:

The angle between a tangent (${AB}$) and a chord (${PS}$) is equal to an angle subtended by the chord ($\angle PRS$).

Since $\angle PRS$ is 60 degrees since it is part of an equilateral triangle, so is angle $ASP$. The argument for angle $BSR$ being 60 degrees is analogous.
 
magneto said:
Given the line ${AB}$ is tangent to the circle at $S$, there is a theorem that states:...

Thanks for replying, magneto but, as far as I can tell, if we know beforehand that $AB$ is a tangent to the circle at $S$, then S certainly lies on AB, and there is nothing to be proved...what do you think?(Smile)
 
anemone said:
I am definitely not saying your approach is wrong, it just that I don't get it, could you elaborate more on that, please? :o

Sure! :)

Do you agree that $M$ is the circumcentre of $PSR$? If so, it can be easily shown that the lines are tangent.

I hope that helps.
 
Pranav said:
Sure! :)

Do you agree that $M$ is the circumcentre of $PSR$? If so, it can be easily shown that the lines are tangent.

I hope that helps.

Yes, I can see QPA and QRB are tangent to the circle which has its center also a circumcenter of triangle PSR, I just don't understand why that implies the angles of ASP, RSB as 60 degree and hence ASB is a straight line. I admit that I don't see how this is obvious for me.(Tmi)
 
anemone said:
Yes, I can see QPA and QRB are tangent to the circle which has its center also a circumcenter of triangle PSR, I just don't understand why that implies the angles of ASP, RSB as 60 degree and hence ASB is a straight line. I admit that I don't see how this is obvious for me.(Tmi)

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(The diagram isn't too accurate, sorry about that. :o )

If we draw a tangent from S, it would form an acute angle $60^{\circ}$ with RS. Since the tangent at S and QRB forms the same angle with RS, they must intersect at the same point.

If you are not satisfied, I am not sure, you can assume that tangent at S intersect PM at B' and then prove BB'=0 or BM=B'M.

I hope it helps. :)
 
Hey Pranav, I see it now.:o Thanks for all the clarification posts and thank you for participating!
 
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