# Thread: Quadrilateral inscribed in a Circle

1. ## Quadrilateral inscribed in a Circle

Hey everyone,

I have a proof here that reads:

A quadrilateral is inscribed in a circle. Prove that the sum of either pair of opposite angles equals two right angles.

Could I get some help as to how to approach this question?

Thanks guys

2. Originally Posted by GreenDay14
Hey everyone,

I have a proof here that reads:

A quadrilateral is inscribed in a circle. Prove that the sum of either pair of opposite angles equals two right angles.

Could I get some help as to how to approach this question?

Thanks guys
Hi GreenDay14,

Locate the circle centre and draw lines from it to all 4 corners of the quadrilateral.
Since each resulting triangle has 2 sides of radius length, they are 4 isosceles triangles.
Now label equal angles at the circumference,
for example A, A, B, B, C, C, D, D.

Then (2A+2B)+(2C+2D)=360 degrees.
(A+B)+(C+D)=180 degrees.
Also (A+D)+(B+C)=180 degrees.

3. Originally Posted by Archie Meade
Hi GreenDay14,

Locate the circle centre and draw lines from it to all 4 corners of the quadrilateral.
Since each resulting triangle has 2 sides of radius length, they are 4 isosceles triangles.
Now label equal angles at the circumference,
for example A, A, B, B, C, C, D, D.

Then (2A+2B)+(2C+2D)=360 degrees.
(A+B)+(C+D)=180 degrees.
Also (A+D)+(B+C)=180 degrees.
I was thinking about doing it this way, but what if the center of the circle is not inside the quadrilateral?

4. Originally Posted by GreenDay14
Hey everyone,

I have a proof here that reads:

A quadrilateral is inscribed in a circle. Prove that the sum of either pair of opposite angles equals two right angles.

Could I get some help as to how to approach this question?

Thanks guys
Which angles are you talking about? The ones between the rectangle edges and the circle? Well, to prove that you have to extend a tangent at the intersection of the 2 shapes. Then, you have a line which has $180^o$. The rectangle takes up $90^o$ (by definition), so the remaining must be $90^o$, which is equal to two right angles.

I'm not sure if that was the angle you wanted, but hope I helped you there. By the way, Green Day is great

5. Originally Posted by icemanfan
I was thinking about doing it this way, but what if the center of the circle is not inside the quadrilateral?
Well spotted icemanfan!

I like your posts quite a lot btw!

We'd have one of the triangles overlapping the other 3.
If the overlapping triangle has equal angles D, which cut across A and C,
this time we'd have [(A-D)+(B+C)]+[(C-D)+(A+B)]=360 degrees.

in both cases, the sum of the opposite angles are A+B+C-D.
Since all 4 angles sum to 360 degrees, the opposite angles sum to 180.

6. Originally Posted by Archie Meade
Well spotted icemanfan!

I like your posts quite a lot btw!
Thanks Archie. Not to rain on your parade, but there is still one more case to consider: one of the sides of the quadrilateral includes the center of the circle. Also, if you are allowed to use the fact that an inscribed angle has half the measure of the central angle, the problem is pretty easy, but I don't know if this is allowed or not.

7. excellent icemanfan!

One of the 4 triangles disappears.
We have 3 isosceles triangles left.
Let's throw out the one with equal angles D.

Then, with the one with equal angles B in the middle,
we have (A+A+B+B+C+C)=360 degrees.
Opposite angles of the quadrilateral are A and (B+C),
(A+B) and C.
The sum in both cases is A+B+C=180 degrees.

All cases ought to be examined.

8. Thanks for all the info guys! A GREAT help as always !

9. I am also instructed to find its converse. I would think I have to use an indirect proof for the converse. But i am a little confused as to how to attempt that. Thanks again for the help guys.

10. Originally Posted by GreenDay14
I am also instructed to find its converse. I would think I have to use an indirect proof for the converse. But i am a little confused as to how to attempt that. Thanks again for the help guys.
What exactly do you mean by the converse?

11. I believe the converse would be that the two are not right triangles? I could be wrong tho...

12. Originally Posted by GreenDay14
I believe the converse would be that the two are not right triangles? I could be wrong tho...
There is no converse to a statement that does not include an implication. The original statement must be of the form "P implies Q" in order to have a converse. The converse of "P implies Q" is "Q implies P."

13. Is that all the question says, GreenDay14? Maybe, by "converse," the question is asking you to prove that if the sum of either pair of opposite angles equals two right angles, the shape inscribed by the circle must be a quadrilateral?

14. Proving the converse will be...

If a quadrilateral's opposite sides sum to 180 degrees,
then it is a cyclic quadrilateral.

There will be alternative ways to show this.
One way will be to examine the cases where an apex of the quadrilateral
does not lie on the circle circumference, examining the angle relationships
for those cases.

15. This is my first posting in this forum, so I don't want to bend anyone out of shape. However, there have been some mistakes made. For example, an "inscribed" quadrilateral must have its four corners on the circle by definition. Therefore the center of the circle MUST be inside the quadrilateral.

The proof is based upon a prior development that an arc [two points on the circle] inscribes and angle at the center that is twice that inscribed on the circumference. Unfortunately, this forum does not allow direct uploading of an image, but only through a link, so I have no image to offer. However, if, from two adjacent corners of the rectangle you draw an angle at the center and one at the circumference, then do the same form the other two points, you could label the center angles as 2A and 2B, and the ones at the circumference [which will appear at the other two points of the quadrilateral] as A and B. it will be seen that 2A + 2B = 360, so A + B = 180 [two right angles.]

If anyone knows of a good place to upload images please let me know. I don't want one of those sites that is filled with ads and pop-ups though please.

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