1) Let ABCDE be a regular pentagon. Prove that AC bisects angle BCE and AD bisects angle BDE.
2) In quadrilateral BCDE, CBD = 31 degrees, CED = 47 degrees, BCE = 85 degrees. Find angle BDE.
When we draw the figure described in the problem, we see that there are vertical angles in the middle of the quadrilaterals.
The Vertical Angles theorem tells us that opposite angles of two intersecting lines are congruent, so angle COB and angle DOE are congruent. (O is the intersection in the middle of the quadrilateral) Since the sum of the angles in a triangle add up to 180 degrees, we see that triangle COB is 85° +31° +x, and triangle DOE is 47° +x+ y (y is angle BDE). Because both are equal to 180° in total, 85° +31°+ x = 47°+x+y.
We simplify to get 116° +x=47° +x+y, 116°-47°=y, Angle BDE = 69°