1. ## Can CaptainBlack help me with this?

Right I have been given the following problem and cannot resolve it. I have had an attempt but without much success. Could anyone help me with this exercise, please?

A cyclic hexagon is a hexagon whose vertices all lie on the circumference of a circle.

The vertices of a cyclic hexagon are labelled in order A to F. Prove that the sum of the interior angles at A, C and E is equal to the sum of the interior angles at B, D and F.

Generalise (concisely) to other cyclic polygons?

2. Originally Posted by Natasha
Right I have been given the following problem and cannot resolve it. I have had an attempt but without much success. Could anyone help me with this exercise, please?

A cyclic hexagon is a hexagon whose vertices all lie on the circumference of a circle.

The vertices of a cyclic hexagon are labelled in order A to F. Prove that the sum of the interior angles at A, C and E is equal to the sum of the interior angles at B, D and F.

Generalise (concisely) to other cyclic polygons?
1. Proving that the sum of the interior angles at verices A, C, and E
equals the sum of the interior angles at vertices B, D, F is equivalent
to proving that a-b+c-d+e-f = 0 where a, b, c, d, e, f are the interior
angles at the vertices A, B, C, D, E, F respectivly.

2. Observe that this property (which I will call the alternating angle
sum property) holds for a regular hexagon, since all of these angles
are equal in a regular hexagon.

3. Observe that for any cyclic hexagon that the alternating angle
sum does not change if we move one vertex along the arc connecting
its neighbours, as long as we leave the vertices in the same realtive order
(sketch of a proof of this is given below).

4. Observe that any cyclic hexagon can be transformed into a regular
hexagon by a sequence of transformations like that described in
para 3 above.

5. Hence conclude that the alternating angle sum property holds
for any cyclic hexagon.

3. Thanks ever so much Captain. That's so clear!

4. How do you proove that

1. We want to prove that a-b+c-d+e-f = 0 where a, b, c, d, e, f
are the interior angles at the vertices A, B, C, D, E, F respectivly.

Do you say that a+b+c+d+e+f = 360 and then what do you do?

5. Originally Posted by Natasha
How do you proove that

1. We want to prove that a-b+c-d+e-f = 0 where a, b, c, d, e, f
are the interior angles at the vertices A, B, C, D, E, F respectivly.

Do you say that a+b+c+d+e+f = 360 and then what do you do?

You are asked to prove that the sum of the interior angles at
vertices A, C and E is equal to the sum of the interior angles
at vertices B, D and F. Using a, b, c, d, e, and f to denote the
interior angles at the corresponding vertices this becomes:

Prove that:

a + c + e = b + d + f,

rearranging, you need to prove that:

a - b + c - d + e - f = 0.

RonL

6. Captain I still don't get this?

I mean the only way I can do this is by saying that:

a+f+b+c+e+d = a+f+b+c+e+d

where angle A is formed of angles f and a, B of a and b, C, b and c, D of c and d, E of d and e and F of e and f.

But I just drew the hexagon and just wrote down the above but don't really prove it???

Could you tell me exactly how you get your first point of

1. We want to prove that a-b+c-d+e-f = 0 where a, b, c, d, e, f
are the interior angles at the vertices A, B, C, D, E, F respectivly.

Don't get it sorry...

7. Originally Posted by Natasha
Captain I still don't get this?

I mean the only way I can do this is by saying that:

a+f+b+c+e+d = a+f+b+c+e+d

where angle A is formed of angles f and a, B of a and b, C, b and c, D of c and d, E of d and e and F of e and f.

But I just drew the hexagon and just wrote down the above but don't really prove it???

Could you tell me exactly how you get your first point of

1. We want to prove that a-b+c-d+e-f = 0 where a, b, c, d, e, f
are the interior angles at the vertices A, B, C, D, E, F respectivly.

Don't get it sorry...

The statement:

"We want to prove that a-b+c-d+e-f = 0 where a, b, c, d, e, f are
the interior angles at the vertices A, B, C, D, E, F respectivly",

is just another way of writting:

"The vertices of a cyclic hexagon are labelled in order A to F.
Prove that the sum of the interior angles at A, C and E is equal
to the sum of the interior angles at B, D and F."

Rewriting:

a-b+c-d+e-f = 0

we have

a+c+e = b+d+f,

which is just another way of saying that "the sum of the interior
angles at A, C and E is equal to the sum of the interior angles
at B, D and F".

This is because a+c+e is the sum of the interior angles at A, C and E,
and b+d+f is the sum of the interior angles at B, D and F.

Hope this helps explain.

RonL

8. The problem here Captain is that you don't seem to be proving anything are you?

I need to use angle sizes I think, no? To state that I have 6 isosceles triangles, no?

9. It's actually cyclic polygons in the question not cyclic hexagons (at the end).

10. Originally Posted by Natasha
It's actually cyclic polygons in the question not cyclic hexagons (at the end).
The only property of the hexagon used in the proof was that
it was cyclic and had an even number of sides (it's not obvious
but it is true).

So the conclusion for a even sided cyclic polygon with vertices
V1, V2, .. Vn is that the sum of the interior angles at V1, V3,..
is equal to the sum of the interior angles at vertices V2, V4, ..

Hope that helps

RonL

11. Thanks captain! but what about an odd sided polygon like a triangle?

12. I received the page proofs for the paper based on this question today

Its to be published in September 2007

RonL

13. Hello, Natasha!

Some basic geometry (plus a little algebra) is required.

A cyclic hexagon is a hexagon whose vertices all lie on the circumference of a circle.

The vertices of a cyclic hexagon are labelled in order A to F.
Prove that the sum of the interior angles at A, C and E
is equal to the sum of the interior angles at B, D and F.
Code:
                A
* o *
* o       o *
F o               o B
*o               o*
o               o
* o               o *
* o       *       o *
* o               o *
o               o
*o               o*
E o               o C
* o       o *
* o *
D

An inscribed angle is measured by one-half its intercepted arc.

. . $\displaystyle \angle A \;=\;\frac{1}{2}\left(\widehat{BC} + \widehat{CD} + \widehat{DE} + \widehat{EF}\right)$

. . $\displaystyle \angle C \;=\;\frac{1}{2}\left(\widehat{DE} + \widehat{EF} + \widehat{FA} + AB\right)$

. . $\displaystyle \angle E \;=\;\frac{1}{2}\left(\widehat{FA} + \widehat{AB} + \widehat{BC} + \widehat{CD}\right)$

Then: .$\displaystyle \angle A + \angle C + \angle E \;=\;\frac{1}{2}\left(2\!\cdot\!\widehat{AB} + 2\!\cdot\!\widehat{BC} + 2\!\cdot\!\widehat{CD} + 2\!\cdot\!\widehat{DE} + 2\!\cdot\!\widehat{EF} + 2\!\cdot\!\widehat{FA}\right)$

Hence: .$\displaystyle \angle A + \angle C + \angle E \;=\;AB + BC + CD + DE + EF + FA \;=\;360^o$

In a similar fashion, it can be shown that: .$\displaystyle \angle B + \angle D + \angle F \;=\;360^o$

14. Originally Posted by Soroban
Hello, Natasha!
I doubt Natasha will be reading this given how old the thread is (though it
would be nice to be proved wrong).

RonL

15. Originally Posted by Soroban
Hello, Natasha!

Some basic geometry (plus a little algebra) is required.

Code:
                A
* o *
* o       o *
F o               o B
*o               o*
o               o
* o               o *
* o       *       o *
* o               o *
o               o
*o               o*
E o               o C
* o       o *
* o *
D
An inscribed angle is measured by one-half its intercepted arc.

. . $\displaystyle \angle A \;=\;\frac{1}{2}\left(\widehat{BC} + \widehat{CD} + \widehat{DE} + \widehat{EF}\right)$

. . $\displaystyle \angle C \;=\;\frac{1}{2}\left(\widehat{DE} + \widehat{EF} + \widehat{FA} + AB\right)$

. . $\displaystyle \angle E \;=\;\frac{1}{2}\left(\widehat{FA} + \widehat{AB} + \widehat{BC} + \widehat{CD}\right)$

Then: .$\displaystyle \angle A + \angle C + \angle E \;=\;\frac{1}{2}\left(2\!\cdot\!\widehat{AB} + 2\!\cdot\!\widehat{BC} + 2\!\cdot\!\widehat{CD} + 2\!\cdot\!\widehat{DE} + 2\!\cdot\!\widehat{EF} + 2\!\cdot\!\widehat{FA}\right)$

Hence: .$\displaystyle \angle A + \angle C + \angle E \;=\;AB + BC + CD + DE + EF + FA \;=\;360^o$

In a similar fashion, it can be shown that: .$\displaystyle \angle B + \angle D + \angle F \;=\;360^o$
The generalisation to a 2n cyclic polygon looks a bit fiddly, though not undoable.

RonL

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