It is known that in △ ABC, be = CE, ∠ DBC = ∠ ACB = 120 °, BD = BC, CD intersection AB is at point e., and de = 3ce is proved
It is proved that & nbsp; ∵ - DBC = - ACB = 120 ° BD = BC ∵ - D = - BCD = 30 ∵ - ACD = 90, BM ⊥ DC over B over M, DM ∵ MC.BM =1/2BC∵AC=1/2BC∴BM=AC∵∠BMC=∠ACM=90∠MEB=∠CEA∴BME≌ACE∴ME=CE=1/2CEDE=3CE
RELATED INFORMATIONS
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