The sum of the interior angles of an n-sided polygon is
SUM = (n-2)∙180°
So for a pentagon, the sum is
SUM = (5-2)∙180° = 3∙180° = 540°
Since all interior angles of a regular pentagon are equal, we
divide that by 5, and get 540°÷5 = 108°
So each of the interior angles of the pentagon measures 108°.
Since ∠B = 108° and ΔABC is isosceles, its base angles are 36° each, and the
same for ΔAED.
Since all three angles at A add to 108°, ∠DAC also = 36°.
Since ΔADC is isosceles, its base angles are equal and 72° each.
Now you have all the angles of ΔADC.
Edwin
