∵ AB = CD, AC = CE,∠ABC =∠CDE=90°
∴由勾股定理,BC² = AC² - AB² = CE² - CD² = DE²
∴ BC = DE
∵ AB = CD,∠ABC =∠CDE = 90°,BC = DE
∴ △ABC全等于△CDE
∴ ∠BAC = ∠DCE,即角A等于角1
∠ACE = 180° - ∠ACB - ∠DCE = 180° - ∠ACB - ∠CAB = 90°
∴ AC⊥CE ■
∠ACB为∠2,∠ECD为∠3
1.
∵AC=CE
∠B=∠D=90°
∴ △ABC全等于△CDE
∴∠A=∠1
2.
∵ △ABC为直角三角形
∴ ∠A+∠2=90°
又∵∠A=∠1
∴∠1+∠2=90°
∵∠1+∠2+∠3=180°
∴∠3=90°
∴AC垂直CE