tangent line to circle Theorem

[logistic_guy]

Prove the theorem by contradiction.



Hint: Assume the line \(\displaystyle m\) is not perpendicular to the the segment \(\displaystyle \overline{QP}\). Then assume the line \(\displaystyle m\) is not tangent to the circle.

 

[topsquark]

Are you going to show some work, or is this just going to be another list of problems to waste our time on?

-Dan

 

[logistic_guy]

Professor Dan is in hereit's an honor. But I don't think that you are serious in helping me.

Are you going to show some work,

Of course. I know that you are here not to help me but I will give you some work because it is very rare to see a reply from you since the exposure of my identity.

Hint: Assume the line \(\displaystyle m\) is not perpendicular to the the segment \(\displaystyle \overline{QP}\).

Here is my attempt. Okay, I assumed that it is not perpendicular, so what is the big deal of this assumption? I also provided a diagram of my assumption to show you how serious I am.

 

[fresh_42]

Non-tangent means secant, which means a proper triangle, which means no right angles.

The rest are technical details.

 

[logistic_guy]

Welcome professor Stefan.

Non-tangent means secant, which means a proper triangle, which means no right angles.

The discussion is now about the non-perpendicular assumption. When I prove it is perpendicular by contradiction, then I use this fact and the second assumption to prove it is tangent (this part should involve the secant).

I used the first assumption which means I assumed that the line \(\displaystyle m\) is not perpendicular to the segment \(\displaystyle \overline{QP}\). I don't see any information that contradict this fact. What about you? Do you see?

 

[logistic_guy]

@topsquark
You could not help me. If you are not good in Geometry, why not just stick with RELATIVITY and DEs?

 

[logistic_guy]

If we assume that the segment \(\displaystyle \overline{QP}\) is not perpendicular to the line \(\displaystyle m\), then there must be a point on \(\displaystyle m\) say \(\displaystyle k\) such that the segment \(\displaystyle \overline{Qk}\) is perpedicular to \(\displaystyle m\).


In a right triangle the segment \(\displaystyle \left(\overline{QP}\right)\) that is opposite to the \(\displaystyle 90^{\circ}\) angle must be the longest leg. It is not the case here. If the segment \(\displaystyle \overline{QP}\) is longer than the segment \(\displaystyle \overline{Qk}\), the segment \(\displaystyle \overline{Qk}\) must lie inside the circle which contradicts our assumption. Then, the segment \(\displaystyle \overline{QP}\) is perpedicular to the line \(\displaystyle m\).

Part one has been proved, but we have not yet proved that the line \(\displaystyle m\) is tangent to the circle. That would be done in the next Episode.

 

[logistic_guy]

If the line \(\displaystyle \textcolor{red}{m}\) is not tangent to the circle, then it must be secant.

Then, we have this geometry:



We have already proved that \(\displaystyle \overline{QP}\) is perpedicular to \(\displaystyle \textcolor{red}{m}\), then the angle \(\displaystyle \angle \text{QPR} = 90^{\circ}\). From the geometry of the triangle we see that the segment \(\displaystyle \overline{QR} = \overline{QP}\), then \(\displaystyle \angle \text{QRP} = \angle\text{QPR} = 90^{\circ}\). This gives the sum of the angles to exceed \(\displaystyle 180^{\circ}\) which contradicts our assumption. Then, line \(\displaystyle \textcolor{red}{m}\) is tangent to the circle.

 

[khansaheb]

Prove the theorem by contradiction.

View attachment 39431

Hint: Assume the line \(\displaystyle m\) is not perpendicular to the the segment \(\displaystyle \overline{QP}\). Then assume the line \(\displaystyle m\) is not tangent to the circle.

We need to assume that

the line m and the circle Q has at least one point in common

 

[Aion]

The only issue I see is that your assumptions in each part of the proof are not entirely clear. You are being asked to prove the following:

A line is tangent to a circle if and only if it is perpendicular to the radius at the point of intersection.

Define [imath]P[/imath]: "A line [imath]L[/imath] is tangent to a circle at point [imath]T[/imath]", and [imath]Q[/imath]: "The line [imath]L[/imath] is perpendicular to the radius at point [imath]T[/imath]".

We now want to prove [imath]P \Rightarrow Q[/imath]. That is, assume [imath]P[/imath] and [imath]\neg Q[/imath], and derive a contradiction.

In words: Assume there exists a line [imath]L[/imath] that is tangent to a circle at point [imath]T[/imath], but is not perpendicular to the radius through [imath]T[/imath].

For the converse direction, [imath]Q \Rightarrow P[/imath], suppose [imath]Q[/imath] and [imath]\neg P[/imath]; that is, assume the line [imath]L[/imath] is perpendicular to the radius at the point of intersection [imath]T[/imath], but that [imath]L[/imath] is not tangent to the circle. This assumption should lead to a contradiction.

 

[logistic_guy]

Define [imath]P[/imath]: "A line [imath]L[/imath] is tangent to a circle at point [imath]T[/imath]",

Your statement here is like: Hey guys I am given line \(\displaystyle m\) that is tangent to the circle at point \(\displaystyle P\), but I will change its name to \(\displaystyle L\) and I will change the point \(\displaystyle P\) to \(\displaystyle T\). Then I will assume that this line is tangent to the circle. You are not assuming anything here because the line is already tangent to the circle. You are only changing its name and its point which will not affect anything.

 

[logistic_guy]

You should write here:

In words: You are given a line [imath]L[/imath] that is tangent to a circle at point [imath]T[/imath], but assume that it is not perpendicular to the radius through [imath]T[/imath].

This would make more sense. But the hint was very clear. It wants you to prove it is perpendicular.

Then, the second proof (will use the first proof) and can be written as:

In words: You are given a line [imath]L[/imath] that is perpendicular to the radius through [imath]T[/imath], but assume that it is not tangent to the circle at [imath]T[/imath].

 

[logistic_guy]

The only issue I see is that your assumptions in each part of the proof are not entirely clear.

Tell me what exactly was not clear? I am not an expert like you in Geometry and Trigonometry but I can enlighten you as I have studied this proof for one month!

 

[logistic_guy]

The only issue I see is that your assumptions in each part of the proof are not entirely clear. You are being asked to prove the following:

A line is tangent to a circle if and only if it is perpendicular to the radius at the point of intersection.

Define [imath]P[/imath]: "A line [imath]L[/imath] is tangent to a circle at point [imath]T[/imath]", and [imath]Q[/imath]: "The line [imath]L[/imath] is perpendicular to the radius at point [imath]T[/imath]".

We now want to prove [imath]P \Rightarrow Q[/imath]. That is, assume [imath]P[/imath] and [imath]\neg Q[/imath], and derive a contradiction.

In words: Assume there exists a line [imath]L[/imath] that is tangent to a circle at point [imath]T[/imath], but is not perpendicular to the radius through [imath]T[/imath].

For the converse direction, [imath]Q \Rightarrow P[/imath], suppose [imath]Q[/imath] and [imath]\neg P[/imath]; that is, assume the line [imath]L[/imath] is perpendicular to the radius at the point of intersection [imath]T[/imath], but that [imath]L[/imath] is not tangent to the circle. This assumption should lead to a contradiction.

I am not against your idea here. It is perfectly fine. Only this should be changed

In words: You are given a line [imath]L[/imath] that is tangent to a circle at point [imath]T[/imath], but assume that it is not perpendicular to the radius through [imath]T[/imath].

You are given the line [imath]L[/imath] to be perpendicular to the radius at the point of intersection [imath]T[/imath], but assume that [imath]L[/imath] is not tangent to the circle.

The idea is that you cannot assume that the Earth has one Moon when it is a fact that it has one Moon. I hope that you understood my point.

 

[Aion]

I am not against your idea here. It is perfectly fine. Only this should be changed

In words: You are given a line [imath]L[/imath] that is tangent to a circle at point [imath]T[/imath], but assume that it is not perpendicular to the radius through [imath]T[/imath].

You are given the line [imath]L[/imath] to be perpendicular to the radius at the point of intersection [imath]T[/imath], but assume that [imath]L[/imath] is not tangent to the circle.

The idea is that you cannot assume that the Earth has one Moon when it is a fact that it has one Moon. I hope that you understood my point.

Thanks for your reply — I see where you're coming from, but I don't fully agree.

According to your interpretation, we're given a line [imath]L[/imath] that is tangent to a circle at point [imath]T[/imath], and then we (incorrectly) assume that it is not perpendicular to the radius through [imath]T[/imath]. But in a proof by contradiction, we're not actually given such a line — we only suppose its existence for the sake of argument, in order to derive a contradiction.

In the case of [imath]P \Rightarrow Q[/imath], we assume [imath]P[/imath] (that [imath]L[/imath] is tangent to the circle at [imath]T[/imath]) and [imath]\neg Q[/imath] (that [imath]L[/imath] is not perpendicular to the radius at [imath]T[/imath]). From this, we deduce that [imath]L[/imath] must intersect the circle at another point — contradicting the assumption that [imath]L[/imath] is a tangent. That contradiction is what justifies [imath]P \Rightarrow Q[/imath].

So in this sense, the line isn't given as a real object with known properties — it's hypothetical, and its properties are assumed only temporarily to show that such a configuration cannot exist.

 

[logistic_guy]

Thanks for your reply — I see where you're coming from, but I don't fully agree.

According to your interpretation, we're given a line [imath]L[/imath] that is tangent to a circle at point [imath]T[/imath], and then we (incorrectly) assume that it is not perpendicular to the radius through [imath]T[/imath]. But in a proof by contradiction, we're not actually given such a line — we only suppose its existence for the sake of argument, in order to derive a contradiction.

In the case of [imath]P \Rightarrow Q[/imath], we assume [imath]P[/imath] (that [imath]L[/imath] is tangent to the circle at [imath]T[/imath]) and [imath]\neg Q[/imath] (that [imath]L[/imath] is not perpendicular to the radius at [imath]T[/imath]). From this, we deduce that [imath]L[/imath] must intersect the circle at another point — contradicting the assumption that [imath]L[/imath] is a tangent. That contradiction is what justifies [imath]P \Rightarrow Q[/imath].

So in this sense, the line isn't given as a real object with known properties — it's hypothetical, and its properties are assumed only temporarily to show that such a configuration cannot exist.

I don't agree with saying we assume [imath]P[/imath] (that [imath]L[/imath] is tangent to the circle at [imath]T[/imath]). You should either say we are given this fact or you assume it is not tangent.

As I said before you cannot assume something when it already does exist. Therefore, we are perfectly agreed that we totally don't agree to each other!

And thanks a lot for the rest of the idea!

 

[Aion]

I don't agree with saying we assume [imath]P[/imath] (that [imath]L[/imath] is tangent to the circle at [imath]T[/imath]). You should either say we are given this fact or you assume it is not tangent.

As I said before you cannot assume something when it already does exist. Therefore, we are perfectly agreed that we totally don't agree to each other!

And thanks a lot for the rest of the idea!

I've been reflecting on your reply and wanted to follow up with one more clarification.

You mentioned that “we cannot assume something that already exists.” But the issue is that in a proof of [imath]P \rightarrow Q[/imath], we're not assuming just one thing — we're temporarily assuming both [imath]P[/imath] and [imath]\lnot Q[/imath] to see if that combination leads to a contradiction. That’s the standard approach in proof by contradiction for implications.

You also said [imath]P[/imath] already exists (is given), so there's no need to assume it. But if [imath]P[/imath] is a tautology (i.e., always true), then [imath]P \rightarrow Q[/imath] becomes logically equivalent to [imath]Q[/imath] itself, and the implication holds trivially — which is clearly not the case here. The implication is meaningful because [imath]P[/imath] isn't always true in general.

From a truth table perspective:

• [imath]P \rightarrow Q[/imath] is false only when [imath]P[/imath] is true and [imath]Q[/imath] is false.
• So to disprove the implication, one would need to find such a case.
• Conversely, to prove the implication, we assume that case (i.e., [imath]P \wedge \lnot Q[/imath]) and derive a contradiction — showing it's not logically possible.

That’s exactly what happens in our geometric proof: we suppose [imath]P[/imath] (that a line [imath]L[/imath] is tangent to the circle at point [imath]T[/imath]), and [imath]\lnot Q[/imath] (that [imath]L[/imath] is not perpendicular to the radius at [imath]T[/imath]). This leads to a contradiction — so the configuration can’t exist, and therefore [imath]P \rightarrow Q[/imath] holds.

In logical terms, the implication [imath]P \rightarrow Q[/imath] is semantically true if there is no case where [imath]P[/imath] is true and [imath]Q[/imath] is false. Our contradiction shows exactly that — assuming [imath]P[/imath] and [imath]\lnot Q[/imath] leads to an impossible scenario. So that one "bad" row in the truth table is ruled out, meaning the implication is valid.

Also, regarding your geometric reasoning: I see that you attempted to depict the situation by assuming [imath]L[/imath] is tangent to the circle and drawing it — but that approach doesn’t incorporate the second part of the assumption, namely that [imath]L[/imath] is not perpendicular to the radius at the point of tangency. Once you try to include both conditions at the same time, the construction clearly breaks down — because such a line simply cannot exist. It violates the very geometry of the circle.

So rather than showing that the assumptions coexist, the attempt at drawing it actually proves our point: assuming both [imath]P[/imath] and [imath]\lnot Q[/imath] leads to an impossibility — which is exactly what the proof is meant to demonstrate.

 

[logistic_guy]

I've been reflecting on your reply and wanted to follow up with one more clarification.

Thank you a lot Aion for being so interested in one of my posts. This just reflects how much you love geometry particularly and proofs generally.

Here is the main issue between your thinking and my thinking: The cold war has divided the world into two parties. Russians and Americans. Let me say it again more precisely \(\displaystyle \rightarrow\) Russians and everyone else!

I am from the Russian school. Since you are from the other school, we will always disagree to each other (even if we agree). This does not mean that you are wrong, it is just that in our school we define things differently!

In the Russian school facts cannot be assumed. So in my school you cannot assume \(\displaystyle P\) \(\displaystyle \left(\right.\)or \(\displaystyle P \rightarrow \lnot Q\)\(\displaystyle \left)\right.\), even if temporarily. We say that you will be given \(\displaystyle P\) and you will assume \(\displaystyle \lnot Q\). Here \(\displaystyle P\) and \(\displaystyle Q\) are facts!

There is an old saying \(\displaystyle \rightarrow\) all roads go to Roma. It just means that if I assumed \(\displaystyle P \rightarrow \lnot Q\) or I was given \(\displaystyle P\) and I assumed \(\displaystyle \lnot Q\), the proof I have given in the posts above will remain the same!

The conclusion is that \(\displaystyle \color{black} \bold{Russians}\) don't agree with what you have said. But don't take it personally as you have given a valid statement in the world of mathematics. And Russians are not mathematicians. Russians are Russians!

FYI. I am not Russian and I don't speak Russian. I just read books written by Russian authors!

 

[Aion]

In the Russian school facts cannot be assumed. So in my school you cannot assume \(\displaystyle P\) \(\displaystyle \left(\right.\)or \(\displaystyle P \rightarrow \lnot Q\)\(\displaystyle \left)\right.\), even if temporarily. We say that you will be given \(\displaystyle P\) and you will assume \(\displaystyle \lnot Q\). Here \(\displaystyle P\) and \(\displaystyle Q\) are facts!

In my opinion, your reasoning is circular. You claim, “we don’t need to assume [imath]P[/imath] because [imath]P[/imath] is a fact.” But unless [imath]P[/imath] has been independently proven or specified as a given axiom, treating [imath]P[/imath] as a “fact” is itself an assumption. So you’ve effectively assumed [imath]P[/imath] in order to avoid assuming [imath]P[/imath], which is circular. This move resembles begging the question, because the point under debate—whether we are allowed to assume [imath]P[/imath]—is being taken for granted.

 

[logistic_guy]

In my opinion, your reasoning is circular. You claim, “we don’t need to assume [imath]P[/imath] because [imath]P[/imath] is a fact.” But unless [imath]P[/imath] has been independently proven or specified as a given axiom, treating [imath]P[/imath] as a “fact” is itself an assumption. So you’ve effectively assumed [imath]P[/imath] in order to avoid assuming [imath]P[/imath], which is circular. This move resembles begging the question, because the point under debate—whether we are allowed to assume [imath]P[/imath]—is being taken for granted.

This has become a paradox. It is either you didn't understand what I said, or you agree with me but contradict yourself at the same time! If I assume that Paris is the capital of the United States, are you sure that my assumption is a fact? If that is a fact, then I am confused!



Give me the name of one book (or one author) that proves this Theorem by contradiction and says we assume \(\displaystyle m\) is tangent to the circle! You have the freedom to exclude the Russian authors.