Question

如图,在△ABC中 ∠ACB=90º,AC=BC ,BE⊥EC,垂足为E,AD⊥EC,AB与BC交于F.

Answer 1

(1)

∵ ∠BCE=90º-∠ACE ∠CAD=90º-∠ACE

∴ ∠BCE=∠CAD

∵ AC=BC ∠ACB=90º

∴ ∠BAC=∠ABC=45º AC=BC=AB/√2

∠FAD=90º -∠AFC=90º -∠ABC-∠BCE=90-45-21=24º

(2)

∵ AC=BC ∠ACB=∠CEB=90º ∠BCE=∠CAD

∴ △ACD ≌ △CBE

CE=AD CD=BE=2cm .

AC=√2/2AB=5√2/2

根据勾股定理

AD =√(AC^2-CD^2)=√[(5/√2)^2-2^2] =√(17/2)=1/2*√34

∴DE=CE-CD=1/2*√34-2