Question

初二数学题······

如图,在△ABC中,AB=AC,角平分线DB、CE交于O.求证:OB=OC

Answer 1

证明:
∵AB=AC
∴∠ABC=∠ACB
∵DB.CE分别平分∠ABC和∠型猜ACB
∴∠OBC=1/2∠ABC ∠OCB=1/2∠ACB
∵∠ABC=∠ACB
∴∠OBC=∠OCB
∴OB=OC
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Answer 2

证腊册明由AB=AC 所轮掘宏以角ABC=角散胡ACB 且DB CE为角平分线 所以角0BC=角0CB 所以 OB=0C