Question

如图,在△ABC中,AB=AC=10CM,BC=16CM,AD⊥BC于D,BE是中线,BE于AD相较于点O,求OD,OE

Answer 1

延长AD至点F,使CF平行于BE,易证△BOD全等于△COF,OD=DF,

又AE=EC,△AOE相似于△AFC,则AO=OF,又AD^2=AC^2-CF^2,AD=6,设OD=x,则DF=x,AO=6-x,6-x=2x,x=2,OD=2.

CF^2=CD^2+DF^2=64+4=68,CF=根号68,OE=CF/2=根号68/2,不会打出根号,就这样了啊