在图①△ABC中,∠ACB=90°,AC=BC,直线MN经过点C,且AD⊥MN于点D,BE⊥MN于E.

1）（1）∠CDA = ∠BEC = 90°
∠DAC = 90 °- ∠DCA = ∠ECB
BC = AC
所以 △ADC≌△CEB
（2）则 BE = CD,CE = AD
所以 AD + BE = CE + CD = DE
2）AD^2 + BE^2 = AC^2
∠BEC = ∠CDA = 90°
∠ECB = 90 °- ∠DCA = ∠DAC
BC = AC
所以 △ADC≌△CEB
则 BE = CD
所以 AD^2 + BE^2 = AD^2 + CD^2 = AC^2