1) Prove that △ADE is all equal to △CDF? AE=CF？ Similarly BF=CE because in Rt△CEF, CE &；; sup2+CF & amp； Sup2= EF & ampsup2, so

AE & ampsup2+BF & amp； sup2= EF & ampsup2？

2) Fold △ACE along CE and △BCF along CF? Because ∠ ace+∠ BCF = ∠ ECF = 45? So after folding,

AC and BC overlap as CP connection PE? PF is available ∠ EPF = ∠ A+∠ B = 90, so AE & sup2+BF & amp； sup2= EP & ampsup2+FP & amp； sup2= EF & ampsup2