Question 503773
in a regualr pentagon ABCDE draw diagonal BE and then find the measure of angle BAE,ABE and BED
<pre>
{{{drawing(400,400,-1.5,1.5,-1.5,1.5,

line(cos(18*pi/180),sin(18*pi/180),cos(90*pi/180),sin(90*pi/180)),
line(cos(90*pi/180),sin(90*pi/180),cos(162*pi/180),sin(162*pi/180)),
line(cos(162*pi/180),sin(162*pi/180),cos(234*pi/180),sin(234*pi/180)),
line(cos(234*pi/180),sin(234*pi/180),cos(306*pi/180),sin(306*pi/180)),
line(cos(306*pi/180),sin(306*pi/180),cos(18*pi/180),sin(18*pi/180)),

green(line(cos(162*pi/180),sin(162*pi/180),cos(18*pi/180),sin(18*pi/180))),

red(arc(0,1,.5,-.5,216,324),arc(cos(162*pi/180),sin(162*pi/180),.6,-.6,0,36),
arc(cos(162*pi/180),sin(162*pi/180),.4,-.4,0,36),

arc(cos(18*pi/180),sin(18*pi/180),.7,-.7,180,252),
arc(cos(18*pi/180),sin(18*pi/180),.77,-.77,180,252),
arc(cos(18*pi/180),sin(18*pi/180),.84,-.84,180,252)

),

locate(cos(18*pi/180)+.05,sin(18*pi/180)+.05,E),
locate(cos(90*pi/180),sin(90*pi/180)+.15,A),
locate(cos(162*pi/180)-.1,sin(162*pi/180),B),
locate(cos(234*pi/180),sin(234*pi/180),C),
locate(cos(306*pi/180),sin(306*pi/180),D) )}}}

The sum of the interior angles of a polygon is given
by the this formula, with N=5

(N-2)×180° = (5-2)×180° = 3×180° = 540°

And since all 5 interior angles of a regular polygon
have equal measure, each interior angle is {{{"540°"/5}}} = 108°

Angle A is an interior angle so the measure of angle A is 108².

Triangle ABE is isosceles, its base angles have sum 180°-108° = 72°
Each base angle is half that, so angle ABE has measure 36°.

Angle AED is an interior angle of the regular pentagon, so
it has measure 108°.  Angle AEB is the other base angle of the
isosceles triangle ABE, so it is 36°.

Angle BED = Angle AED - Angle AEB = 108°-36° = 72°  

Edwin</pre>