# Nonagon Problem

• Sep 23rd 2010, 06:05 AM
PythagorasNeophyte
Nonagon Problem
I have a question about polygons. Here's how it goes:

Attachment 19017

I know that the sum of interior angles of the regular nonagon is 180º(n-2)=
180n -360
= 180(9) -360
=1260º

So each of the interior angle is 1260º/9 = 140º

angle PCD + angle BCD = 180º (adjacent angles on a straight line)

angle PCD = 180º - 140º

angle PCD = 40º

So that explains part a).

Can you please give me tips on part b)?
• Sep 23rd 2010, 07:35 AM
Traveller
Here are some hints :

i) Triangle DCE is isosceles.

ii) You will get this from the previous part.

iii) Triangle BEP is isosceles.
• Sep 24th 2010, 02:35 AM
PythagorasNeophyte
Hello Traveller.

Thank you for your reply, but what are the proofs that triangle DCE and trianglle BEP are isosceles triangles?
• Sep 24th 2010, 02:48 AM
Educated
This is a regular nonagon, which means that all side lengths are the same.

Triangle DCE uses 2 lengths of the regular nonagon, therefore it must be an iscoceles triangle.

Triangle BEP are isoceles because length BP and EP both use extended lines of the regular nonagon sides. These lines that meet have to be equal, because the angles of the exterior are the same.

And also all regular polygons have symmetry, so if you drew a line from one corner to the other, the angles would be the same, thus making it an iscoceles triangle.
• Sep 24th 2010, 03:14 AM
PythagorasNeophyte
Thank you so much Educated!

So the answer for b)i) will be:

$CD = DE$ (sides of isos. triangle)

$Angle CDE is 140 °$ (one of the int. angles of nonagon)

$Angle CDE + angle DCE + angle DEC = 180 °$ (angle sum of triangle)

$Angle DCE + angle DEC = 40 °$

$Angle DCE = 20 °$

----------
As for part ii):

Angle BCD is $140 °$.
Angle DCE is $20 °$.

Angle BCE is $140 °-20 °=120 °$

------------
As for part iii):

Angle BEF + Angle DEB = angle DEF

So
$Angle BEF = 140 °-2(20 °)
=100 °$

Problem solved.