Question

如图，在△ABC，AB=AC，EF交AB于E，交AC的延长线于F，交BC于D，且BE=CF，求证：DE=DF

Answer 1

证：过E做EP平行AC

  ∵AB=AC

  ∴∠EBP=∠ACB=∠EPB

  ∴BE=EP=CF（BE=CF） ∠EPD=180°-∠EPB，∠DCF=180°-∠ACB

  ∴∠EPD=∠DCF

  ∴△EPD≌△FCD

  ∴DE=DF

Answer 2

过E做EH平行AD;交BC于H;
△ABC中,AB=AC

∴∠ABC=∠ACB;
∵EH//AF;
∴∠EHB=∠ACB=∠ABC;
∴BE=EH=CF;
EH//CF;
∴∠FEH=∠DFC;∠EDH=∠CDF
∴△EDH≌△DCF;
∴EF=DF;