以C为中心,将△CAE顺时针旋转90°,
使得A与D重合,E到F。
∵△CAE≌△CDF,
∴∠ECF=90°,连EF,
由CE=CF=2a,∠CFE=45°,∴EF=(2√2)a,
由DF=AE=a,DE=3a,
有EF²+DF²=DE²=9a²,△EFD是直角三角形,
∴∠EFD=90°,即∠CFD=135°,
由余弦定理:
CD²=DF²+CF²-2×DF×CFcos135°
=a²+4a²+2×a×2a×√2/2
=5a²+2a²√2
∴CD=a√(5+2√2)
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