Question

如图（1）已知△ABC是等边三角形,点D、E、F分别在边AB、BC、AC上,且∠1=∠2=∠3

Answer 1

DP与AC交于O,

∠ADP+∠1=60°

∠AFP+∠2=60°∠1=∠2

∠ADP=∠AFP

∠AOD=∠FOP

ΔAOD∽ΔPOF

AO/OD=PO/OF

∠AOP=∠FOD

ΔAOP∽ΔDOF

∠OAP=∠ODF=60°

∠BAP=120°∠BAP+∠B=180°

AP//BC

解法2：

如图,过DF中点G做BC平行线,交AC于H,可知H为AC的中点.

AJ=EI

EI=PK

AJ=PK

AP//GH//BC