In the pentagonal ABCD, ab = CD = de = BC + AE = 2, angle B = angle e = 90 degrees, find the area of the pentagonal ABCDE

In the pentagonal ABCD, ab = CD = de = BC + AE = 2, angle B = angle e = 90 degrees, find the area of the pentagonal ABCDE

Extend CB to e ', such that e'b = AE ≌ e = ≌ Abe', ab = de ≌ EDA ≌ Abe ≌ ad = AE ', CE' = BC + be '= BC + AE = CD = 2, and AC = AC ≌ ACD ≌ ace', so s Pentagon ABCDE = s △ ABC + s △ ade + s △ ACD = s △ ABC + s △ Abe '+ s △ ACD = s △ ace' + s △ ACD = 2S △ ace '= 2 * 1 / 2 * ab

It is known that: as shown in the figure, in △ ABC, ∠ C = 90 °, cm ⊥ AB in M, at bisection ∠ BAC intersects cm in D, intersects BC in T, de ∥ AB through D intersects BC in E, verification: CT = be

It is proved that: TF ⊥ AB is made in F through t, ∵ at bisection ⊥ BAC, ∵ ACB = 90 °, CT = TF (the distance from the point on the bisection line to both sides of the angle is equal), ∵ ACB = 90 °, cm ⊥ AB, ∵ ADM + ⊥ dam = 90 °, ATC + ⊥ cat = 90 °, ∵ at bisection ⊥ BAC, ∵ dam = ⊥ cat, ∵ ADM = ⊥ ATC

It is known that: as shown in the figure, in △ ABC, ∠ C = 90 °, cm ⊥ AB in M, at bisection ∠ BAC intersects cm in D, intersects BC in T, de ∥ AB through D intersects BC in E, verification: CT = be

It is proved that: TF ⊥ AB is made in F through t, ∵ at bisection ⊥ BAC, ∵ ACB = 90 °, CT = TF (the distance from the point on the bisection line to both sides of the angle is equal), ∵ ACB = 90 °, cm ⊥ AB, ∵ ADM + ⊥ dam = 90 °, ATC + ⊥ cat = 90 °, ∵ at bisection ⊥ BAC, ∵ dam = ⊥ cat, ∵ ADM = ⊥ ATC