In the Pentagon ABCDE, ab = BC = AE = de In the Pentagon ABCDE, ab = BC = AE = De, BAE = b = BCD = D = e prove CD = de The answer to "wj1232944" is very speechless. Look at the question carefully. Only four sides are equal Now that you know the answer, add guides

In the Pentagon ABCDE, ab = BC = AE = de In the Pentagon ABCDE, ab = BC = AE = De, BAE = b = BCD = D = e prove CD = de The answer to "wj1232944" is very speechless. Look at the question carefully. Only four sides are equal Now that you know the answer, add guides

Because AB = BC = AE = de angle BAE = angle B = angle BCD = angle d = angle e
So this is a regular pentagon
So CD = de

In the pentagonal ABCDE, AE ⊥ ed ab ∥ CD ∠ a = ∠ B = ∠ D, find ∠ C

AE⊥ED
Then ∠ e = 90
AB//CD
Then, B + C = 180
Let a = - B = - D = X
Then the sum of the inner angles of the Pentagon
∠A+∠B+∠C+∠D+∠E=540
That is: x + X + (180-x) + X + 90 = 540
2x=270
x=135
be
∠C=180-x=180-135=45