$证明:(1)设AB=BD=DE=EC=m，则AD=\sqrt{2}m,CD=2m,AE=\sqrt{5}m,AC=\sqrt{10}m$ $∴\frac{AD}{CD}=\frac{\sqrt{2}m}{2m}=\frac{\sqrt{2}}{2},\frac{DE}{DA}=\frac{m}{\sqrt {2}m}=\frac{\sqrt{2}}{2}$ $\frac{AE}{CA}=\frac{\sqrt{5}m}{\sqrt{10}m}=\frac{\sqrt{2}}{2},∴\frac{AD}{CD}=\frac{DE}{DA}=\frac{AE}{CA},∴△ADE∽△CDA$ $(2)由(1)，知△ADE∽△CDA,∴∠DAE=∠3$ $∵∠B=90°，AB=BD,∴∠1=45°$ $∵∠1=∠2+∠DAE,∴∠2+∠3=∠1=45°,∴∠1+∠2+∠3=90°$ $证明:(1)由勾股定理得AD=\sqrt{2^{2}+1^{2}}=\sqrt{5},DE= \sqrt{3^{2}+1^{2}}=\sqrt{10}$ $AB=\sqrt{1^{2}+1^{2}}=\sqrt{2},AC= \sqrt{1^{2}+3^{2}}=\sqrt{10}$ $∵AE=5,BC=2,∴\frac{AD}{AB}=\frac{\sqrt{10}}{2},\frac{DE}{BC}=\frac{\sqrt{10}}{2},\frac{AE}{AC}=\frac{\sqrt{10}}{2}$ $∴\frac{AD}{AB}=\frac{DE}{BC}=\frac{AE}{AC},∴△ADE∽△ABC$ $(2)由(1)，得△ADE∽△ABC,∴∠ADE=∠ABC=90°+45°=135°$ $∴∠1+∠2=180°-135°=45°$ $证明:(1)由勾股定理 AB=2\sqrt{5},AC=\sqrt{5},BC=5$ $∴易得AB^{2}+AC^{2}=BC^{2},∴ △ABC 为直角三角形$ $(2)△ABC和△DEF相似,理由:$ $由勾股定理，得DE=4\sqrt{2},DF=2\sqrt{2},EF=2\sqrt{10}$ $∴\frac{AB}{DE}=\frac{2\sqrt{5}}{4\sqrt{2}}=\frac{\sqrt{10}}{4},\frac{AC}{DF}=\frac{\sqrt{5}}{2\sqrt{2}}=\frac{\sqrt{10}}{4},\frac{BC}{EF}=\frac{5}{2\sqrt{10}}=\frac{\sqrt{10}}{4}$ $∴\frac{AB}{DE}=\frac{AC}{DF}=\frac{BC}{EF}$ $∴△ABC∽△DEF$
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