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.
+128475 2.
B C
31 85
E 47 D
Let BD and CE intersect at P
We have triangles BPC and EPD
Angle CDB = angle CBP = 31
Angle BCE = angle BCP = 85
So angle BPC = 180 - 31 - 85 = 64 = angle EPD
So in triangle EPD
Angle CED = angle PED
Angle BDE = 180 - angle EPD -angle PED
Angle BDE = 180 - 64 - 47 = 69°
