(1)
∵CE=CD AC=BC ∠CAE=∠
ADC+∠ACD=∠ABC+∠ABD=∠CBD(三角形外角及圆周角定理)
∴△ACE ≌△BCD
AE=BD
(2)
∵△ACE ≌△BCD
AE=BD ∠ACE=∠BCD
AD+BD=AD+AE=DE=√2CD
DE^2=2CD^2=CD^2+CE^2(符合勾股定理边与边的关系)
∠DCE=90(DE对边为直角)
∠DCE=∠ACD+∠ACE=∠ACD+∠BCD=90
∴AC⊥BC
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