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Joke abcbcbcbcbd

Solution: solution: AE⊥CD, AE=CD,

Reason: extend AE to CD point p,

∵ in △ABE and △CBD,

AB=BC∠ABE=∠DBCBE=BD,

∴△ABE≌△CBD(SAS),

∴AE=DC,∠AEB=∠CDB,∠DCB=∠EAB,

∠∠EAB+∠AEB = 90,

∴∠AEB+∠DCB=90,

∠∠AEB =∠CEP,

∴∠BCD+∠CEP=90,

∴AE⊥CD.