征解问题[17] 平面几何(已解决)

如图,已知等边\(\triangle ABC\),\(\angle ABD=20^\circ\),\(E\)是\(BD\)的中点,\(\angle CBD\)的平分线交\(CE\)于点\(F\),连接\(DF\).求证:\(\angle BFD=\angle ADF\).
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证明    如图,延长\(BF\)交\(AC\)于点\(G\),欲证\(\angle BFD=\angle ADF\),须证\(\angle DFG=\angle FDG\),即求证\(DG=FG\).

2过点\(F\)作\(HI\parallel BD\),分别交\(BC,AC\)于点\(H,I\).
由\(BF\)平分\(\angle DBC\),可证\(BH=HF\);
由题意可得\(\angle BDG=\angle BGD=\angle FIG\),易证\(FI=FG=FH\);
即\(FI=FG=FH\).
则连接\(HG\),可得\(HG\perp AC\).
3所以\(HC=2GC=AD+GC\),可得\(DG=BH=FG\),
即可证\(\angle BFD=\angle ADF\).

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