
1 Attachment(s)
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(n2)=
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)?

Here are some hints :
i) Triangle DCE is isosceles.
ii) You will get this from the previous part.
iii) Triangle BEP is isosceles.

Hello Traveller.
Thank you for your reply, but what are the proofs that triangle DCE and trianglle BEP are isosceles triangles?

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.

Thank you so much Educated!
So the answer for b)i) will be:
$\displaystyle CD = DE$ (sides of isos. triangle)
$\displaystyle Angle CDE is 140 °$ (one of the int. angles of nonagon)
$\displaystyle Angle CDE + angle DCE + angle DEC = 180 °$ (angle sum of triangle)
$\displaystyle Angle DCE + angle DEC = 40 °$
$\displaystyle Angle DCE = 20 °$

As for part ii):
Angle BCD is $\displaystyle 140 °$.
Angle DCE is $\displaystyle 20 °$.
Angle BCE is $\displaystyle 140 °20 °=120 °$

As for part iii):
Angle BEF + Angle DEB = angle DEF
So
$\displaystyle Angle BEF = 140 °2(20 °)
=100 °$
Problem solved.