tangent line to circle Theorem

[Aion]

Thank you for your reply. I think we’re converging on the core of our disagreement: how we treat assumptions versus facts, and how they function in a proof.

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!'

That’s actually the point I was trying to make. If you assume something (like Paris is the U.S. capital), you're not saying it’s a fact—you’re creating a hypothetical scenario. The truth or falsity of that assumption doesn’t matter at first; we analyze its consequences. If assuming Paris is the capital leads to a contradiction (e.g., Washington, D.C. is already the capital), then the assumption must be false.

That’s exactly how proof by contradiction works:

1. Assume something (e.g., a line is tangent to a circle),
2. Follow its logical consequences,
3. If it leads to a contradiction, then the assumption is false.

So again, an assumption is not a fact—it's a tool. I think we're just using the term "assume" in different ways.

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.

Thank you for your challenge. You asked for a reference (non-Russian) where the theorem is proved by contradiction, assuming that a line is tangent to a circle.

Here is one:

Book: Inledande geometri för högskolestudier by Stefan Diehl,
Theorem 21 (page 43): A tangent to a circle is perpendicular to the radius at the point of contact.

Theorem 21.
A tangent to a circle is perpendicular to the radius at the point of tangency.


Proof (by contradiction):
Assume there exists a tangent line [imath]l[/imath] that is not perpendicular to the radius drawn from the center [imath]M[/imath] to the point of tangency [imath]P[/imath].

Draw the perpendicular (the normal) from [imath]M[/imath] to [imath]l[/imath]; let this line intersect [imath]l[/imath] at a point [imath]A[/imath], where [imath]A \ne P[/imath].

Now, mark a point [imath]B \ne P[/imath] on [imath]l[/imath] such that [imath]PA = AB[/imath], and draw the segment [imath]MB[/imath].

We now consider triangles [imath]\triangle AMP[/imath] and [imath]\triangle AMB[/imath]:

• [imath]PA = AB[/imath] (by construction),
• [imath]\angle PAM = \angle BAM = 90^\circ[/imath] (since [imath]AM[/imath] is the normal to [imath]l[/imath]),
• [imath]AM[/imath] is a common side.

Therefore, by the SAS criterion ([imath]\text{Side–Angle–Side}[/imath]), the two triangles are congruent:
[imath]\triangle AMP \cong \triangle AMB[/imath].

Hence, [imath]MP = MB[/imath], so [imath]B[/imath] also lies on the circle (since its distance to the center is equal to the radius).

But this contradicts the assumption that [imath]l[/imath] is a tangent, since a tangent intersects a circle in exactly one point.

[imath]\Rightarrow[/imath] Contradiction. Therefore, our initial assumption was false.

Thus, the tangent [imath]l[/imath] must be perpendicular to the radius at point [imath]P[/imath].

 

[logistic_guy]

Thank you for your reply. I think we’re converging on the core of our disagreement: how we treat assumptions versus facts, and how they function in a proof.

Thank you too Aion. Before I reply to what you have written, I need to know three things about you.

\(\displaystyle \bold{1.}\) Is English your native language?
\(\displaystyle \bold{2.}\) What is your math level?
\(\displaystyle \bold{3.}\) What is your field?

If you think that any of these questions is personal, you don't have to answer it.

 

[Aion]

Thank you too Aion. Before I reply to what you have written, I need to know three things about you.

\(\displaystyle \bold{1.}\) Is English your native language?
\(\displaystyle \bold{2.}\) What is your math level?
\(\displaystyle \bold{3.}\) What is your field?

If you think that any of these questions is personal, you don't have to answer it.

I appreciate your interest in understanding my background—though I believe it's most productive to keep the focus on the mathematics and the reasoning itself. After all, the strength of a proof lies in its logic, not in who presents it.

That said, I'm happy to share a bit about myself:

1. English is not my native language, but I use it regularly.
2. I’ve taken a university-level course in logic, though it was about three years ago now.
3. The topic at hand falls under elementary mathematics, so it’s well within the scope of what I’m comfortable discussing.

The proof I shared earlier is a direct translation from a Swedish textbook: Inledande geometri för högskolestudier by Stefan Diehl. It appears on page 43 as Theorem 21 and gives a clear proof by contradiction, starting with the assumption that a line is tangent to a circle but not perpendicular to the radius. The contradiction arises when this leads to two points of intersection, contradicting the definition of a tangent.

If there's any specific part you'd like to challenge or explore further, I’d be glad to continue the discussion.

 

[logistic_guy]

Book: Inledande geometri för högskolestudier by Stefan Diehl,
Theorem 21 (page 43): A tangent to a circle is perpendicular to the radius at the point of contact.

Theorem 21.
A tangent to a circle is perpendicular to the radius at the point of tangency.

Proof (by contradiction):

Assume there exists a tangent line [imath]l[/imath] that is not perpendicular to the radius drawn from the center [imath]M[/imath] to the point of tangency [imath]P[/imath].

First time to know and see that there are some authors assume facts to contradict something! If I didn't see this now, I would never believe you!

So again, an assumption is not a fact—it's a tool. I think we're just using the term "assume" in different ways.

This statement will make us go back to square one. I understand your point, so this time I will let it go!

English is not my native language, but I use it regularly.

I asked you about your language because I have noticed that your English is well organized. Now after telling me that it is not your native language, I find it very difficult to believe that you are able to write like this without any help.

If there's any specific part you'd like to challenge or explore further, I’d be glad to continue the discussion.

If no body is helping you to write and give this information (other than reading books and pages to gain knowledge), then you have talent in English as well as in geometry and trigonometry. If everything you have said in this thread (since post #\(\displaystyle 10\)) was purely human, then you win and I lose.

And I hope to see your talent in my future Geometry problems (if there will be any)! Thanks a lot Aion for the new information you gave me today. As @Dr.Peterson once said, everyday we learn something \(\displaystyle \textcolor{red}{\bold{NEW}}\).

 

[logistic_guy]

Proof (by contradiction):
Assume there exists a tangent line [imath]l[/imath] that is not perpendicular to the radius drawn from the center [imath]M[/imath] to the point of tangency [imath]P[/imath].

Draw the perpendicular (the normal) from [imath]M[/imath] to [imath]l[/imath]; let this line intersect [imath]l[/imath] at a point [imath]A[/imath], where [imath]A \ne P[/imath].

Now, mark a point [imath]B \ne P[/imath] on [imath]l[/imath] such that [imath]PA = AB[/imath], and draw the segment [imath]MB[/imath].

We now consider triangles [imath]\triangle AMP[/imath] and [imath]\triangle AMB[/imath]:
\(\displaystyle \textcolor{purple}{\bold{*********}}\)

Hey @Aion, I tried to visualize this idea but I failed. Can you please provide a sketch to see where I made a mistake?

 

[Aion]

Proof (by contradiction):
Assume there exists a tangent line [imath]l[/imath] that is not perpendicular to the radius drawn from the center [imath]M[/imath] to the point of tangency [imath]P[/imath].

Draw the perpendicular (the normal) from [imath]M[/imath] to [imath]l[/imath]; let this line intersect [imath]l[/imath] at a point [imath]A[/imath], where [imath]A \ne P[/imath].

Now, mark a point [imath]B \ne P[/imath] on [imath]l[/imath] such that [imath]PA = AB[/imath], and draw the segment [imath]MB[/imath].

We now consider triangles [imath]\triangle AMP[/imath] and [imath]\triangle AMB[/imath]:
\(\displaystyle \textcolor{purple}{\bold{*********}}\)

Hey @Aion, I tried to visualize this idea but I failed. Can you please provide a sketch to see where I made a mistake?

Here’s the figure from the book that the proof is based on. It shows the setup with the center [imath]M[/imath], the tangent line [imath]\ell[/imath], the point of tangency [imath]P[/imath], the foot of the perpendicular [imath]A[/imath], and the reflected point [imath]B[/imath]. You can see how triangles [imath]\triangle AMP[/imath] and [imath]\triangle AMB[/imath] relate to each other in the configuration. Let me know if this helps.

 

[logistic_guy]

View attachment 39623

Here’s the figure from the book that the proof is based on. It shows the setup with the center [imath]M[/imath], the tangent line [imath]\ell[/imath], the point of tangency [imath]P[/imath], the foot of the perpendicular [imath]A[/imath], and the reflected point [imath]B[/imath]. You can see how triangles [imath]\triangle AMP[/imath] and [imath]\triangle AMB[/imath] relate to each other in the configuration. Let me know if this helps.

He told us to assume \(\displaystyle l\) as a tangent line but he drew it as a secant? I am confused!

 

[khansaheb]

Euclid's theorem regarding the angle between a tangent and a radius of a circle states that a tangent to a circle is perpendicular to the radius drawn to the point of contact. This theorem is also mentioned in Euclid's Elements, specifically in Book III, Proposition 18, where it is stated that

if a straight line touches a circle, and a straight line is drawn from the center to the point of contact, then the straight line so drawn will be perpendicular to the tangent.
This theorem was taught to me in eighth grade.

 

[logistic_guy]

This theorem was taught to me in eighth grade.

And how old are you now?

 

[jonah2.0]

Beer drenched non sequitur ramblings follow.

And how old are you now?

Doing jumping jacks naked is not a pleasant activity.
It's still better done wearing briefs and shorts.

 

[Aion]

He told us to assume \(\displaystyle l\) as a tangent line but he drew it as a secant? I am confused!

Great observation — and that’s actually the heart of the contradiction! We start by assuming that line [imath]\ell[/imath] is a tangent, but not perpendicular to the radius. Then, by constructing point [imath]B[/imath] and showing that it also lies on the circle, we end up with a line that intersects the circle at two points — which is the definition of a secant, not a tangent. That contradiction tells us our original assumption must be false. So the only way for [imath]\ell[/imath] to truly be a tangent is if it’s perpendicular to the radius at the point of contact. In short: the sketch looks like a secant because the assumption leads us to a contradiction — and that’s exactly what we want in a proof by contradiction.