Cyclic Quad: How to prove the figure in the graphic is a cyclic quadrilateral?

[urimagic]

Hi friends,

I trust you are all well and in good shape. In the attached drawing we must prove that OTBG is a cyclic quad. I will give the explanation, however I fail to fathom how their explanation proves anything. Just a note, the indicators that show OBG and OTG to be 90deg, ...I put them in, they were not given.

So the only truth we can get from this drawing is that OTG = 90deg, because of the midpoint rule, and OBG also is 90deg, because of the center line to tangent rule. This is also how the explanation is depicted on the memo...HOWEVER, it is now accepted that OTBG is a cyclic quad, because a line subtends equal angles....I presume the line referred to here is line OG. But surely this is not prove enough that TOG + TBG = 180deg's ?...Could someone explain this to me, please?

 

[Dr.Peterson]

In the attached drawing we must prove that OTBG is a cyclic quad. I will give the explanation, however I fail to fathom how their explanation proves anything. Just a note, the indicators that show OBG and OTG to be 90deg, ...I put them in, they were not given.

So the only truth we can get from this drawing is that OTG = 90deg, because of the midpoint rule, and OBG also is 90deg, because of the center line to tangent rule. This is also how the explanation is depicted on the memo...HOWEVER, it is now accepted that OTBG is a cyclic quad, because a line subtends equal angles....I presume the line referred to here is line OG. But surely this is not prove enough that TOG + TBG = 180deg's ?...Could someone explain this to me, please?View attachment 35681

This follows from the converse of Thales' Theorem: Each of B and T lies on the circle whose center is the midpoint of OG. Therefore, all four of these points lie on the same circle.

It would be helpful if you would show the entire problem and the entire proof you are trying to understand, as context (and exact wording) is often important. But I think what I have said is correct, based on what you have said.

 

[urimagic]

Hi Dr.Peterson,

I really do appreciate your assistance with helping me understand all this. Attached the actual question and answer. We have to prove that OTBG is a cyclic quad, and ONLY if it is, will all the angles, OTB and G be lying on the circle, however we must first prove this is indeed a cyclic quad...so, all they say is that OTG = 90deg, because of the midpoint rule, and OBG also is 90deg, because of the center line to tangent rule. And now this proves that OTBG is a cyclic quad....I really cannot see how those two angles, being both 90deg, is enough to prove that OTBG is a cyclic quad...

 

[Dr.Peterson]

We have to prove that OTBG is a cyclic quad, and ONLY if it is, will all the angles, OTB and G be lying on the circle, however we must first prove this is indeed a cyclic quad.

This is the definition of a cyclic quadrilateral, so I'm not sure why you used so many words here!

so, all they say is that OTG = 90deg, because of the midpoint rule, and OBG also is 90deg, because of the center line to tangent rule. And now this proves that OTBG is a cyclic quad....I really cannot see how those two angles, being both 90deg, is enough to prove that OTBG is a cyclic quad...

It would indeed be good if they stated a reason, since they asked you to give one!

But did the answer I gave not help at all??

Please show what theorems you have learned about angles or triangles inscribed in circles or semicircles, so we can see that to call the one you need.

I said this:

This follows from the converse of Thales' Theorem: Each of B and T lies on the circle whose center is the midpoint of OG. Therefore, all four of these points lie on the same circle.

Here are some sources:

Thales's theorem - Wikipedia

en.wikipedia.org

Thales theorem - Math Open Reference

www.mathopenref.com

https://mathvox.com/geometry/triangles/chapter-5-right-triangle/converse-to-thales-theorem-in-a-circle/

 

[urimagic]

Dr.Peterson,

Okay, I have decided to look at this differently....Something I would like to put on the table quick is to say that I am familiar with basically all theorems regarding circles, etc....

What I now did was to actually draw a circle going through all the said points, and I saw the theorem present in the new circle that any angle on the circumference of the circle, subtended by the diameter of a circle, will be 90deg. So, this leads me to believe that OG MUST be the diameter of the new circle, but still I cannot see how this proves OTBG be a cyclic quad...OTG is the same angle top and bottom circles, but OBG in top circle, is not the same as TBG in the bottom circle, and it is TBG that actually is a ligit angle of the cyclic quad, and not OBG...I fail to see how OTG and OBG, being 90 deg each, actually proof that OTBG is a cyclic quad....Maybe I'm also failing in explaining my problem here....

 

[urimagic]

This is the definition of a cyclic quadrilateral, so I'm not sure why you used so many words here!


It would indeed be good if they stated a reason, since they asked you to give one!

But did the answer I gave not help at all??

Please show what theorems you have learned about angles or triangles inscribed in circles or semicircles, so we can see that to call the one you need.

I said this:

Here are some sources:

Thales's theorem - Wikipedia

en.wikipedia.org

Thales theorem - Math Open Reference

www.mathopenref.com

https://mathvox.com/geometry/triangles/chapter-5-right-triangle/converse-to-thales-theorem-in-a-circle/

Dr.Peterson,
I have already posted, but I think I have it now.... The mere fact that we have two 90deg angles on the circumference of the circle, being subtended by the same line, OG, which has to be the diameter, means that all 4 points are actually lying on the circle, and therefore must be a quad....I'm right with this? I think my problem was that line OB threw me off...

 

[blamocur]

I am familiar with basically all theorems regarding circles, etc....

Apparently not all of them But here is another way to approach this problem: consider an arbitrary right triangle ABC with AC being the hypotenuse, and let D be the middle of AC (i.e. AD = DC); what can you tell about the BD distance?

 

[Dr.Peterson]

Dr.Peterson,
I have already posted, but I think I have it now.... The mere fact that we have two 90deg angles on the circumference of the circle, being subtended by the same line, OG, which has to be the diameter, means that all 4 points are actually lying on the circle, and therefore must be a quad....I'm right with this? I think my problem was that line OB threw me off...

Yes, I think you understand.

I assume you meant, "all 4 points are actually lying on the circle, and therefore must be a cyclic quad", because, of course, this is the definition.

I'm not sure why you missed this for so long, but I suspect it is that you were distracted, as you say, by the presence of other lines. You need to focus on definitions: As soon as you see that the four vertices are on a circle, as in both of these last drawings, you should have understood. It doesn't matter that two diagonals of the quadrilateral are also shown.

It's also possible that you focused too much on some particular theorem about the angles in a cyclic quadrilateral, as when you said,

OTG is the same angle top and bottom circles, but OBG in top circle, is not the same as TBG in the bottom circle, and it is TBG that actually is a ligit angle of the cyclic quad, and not OBG

It is not clear what you mean by "top and bottom circles".

But apparently you are beyond that confusion now.

 

[urimagic]

Apparently not all of them But here is another way to approach this problem: consider an arbitrary right triangle ABC with AC being the hypotenuse, and let D be the middle of AC (i.e. AD = DC); what can you tell about the BD distance?

BD would be equal to DC, and equal to AD, making ABD and BCD two isosceles triangles..

 

[urimagic]

Yes, I think you understand.

I assume you meant, "all 4 points are actually lying on the circle, and therefore must be a cyclic quad", because, of course, this is the definition.

I'm not sure why you missed this for so long, but I suspect it is that you were distracted, as you say, by the presence of other lines. You need to focus on definitions: As soon as you see that the four vertices are on a circle, as in both of these last drawings, you should have understood. It doesn't matter that two diagonals of the quadrilateral are also shown.

It's also possible that you focused too much on some particular theorem about the angles in a cyclic quadrilateral, as when you said,

It is not clear what you mean by "top and bottom circles".

But apparently you are beyond that confusion now.

Thank you Dr.Peterson....I do appreciate..