DF is the perpendicular bisector od BC , DE=BF=CF. Both DE and CB are perpendicular to AB
(a) prove DF //AB
(b)prove BD is a median of Δ ABC
maths (triangle) 2?
回答 (1)
2016-12-02 12:22 pm
✔ 最佳答案
Sol(a)
∠DFC=90度=∠EBF
DF//AB
(b)
DE//AB
∠EAD=∠FDC
∠AED=90度=∠DFC
DE=FC
△AED全等於△DFC (AAS)
AD=CD
BD為△ABC之一中線
收錄日期: 2021-04-30 21:57:12
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