- #1
Natasha1
- 494
- 9
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?
My answer so far
I drew the hexagon and labelled 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.
And wrote from the drawing we can see that
a+f+b+c+e+d = a+f+b+c+e+d
But I don't think this is really a clear way of proving is it? Do I need to use angles dimensions and sides?
Where do I go from here basically?
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?
My answer so far
I drew the hexagon and labelled 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.
And wrote from the drawing we can see that
a+f+b+c+e+d = a+f+b+c+e+d
But I don't think this is really a clear way of proving is it? Do I need to use angles dimensions and sides?
Where do I go from here basically?