Here are some hints :
i) Triangle DCE is isosceles.
ii) You will get this from the previous part.
iii) Triangle BEP is isosceles.
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)?
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:
(sides of isos. triangle)
(one of the int. angles of nonagon)
(angle sum of triangle)
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As for part ii):
Angle BCD is .
Angle DCE is .
Angle BCE is
------------
As for part iii):
Angle BEF + Angle DEB = angle DEF
So
Problem solved.