如题所述
âµâ BFD=90°=â BCD
â´AãBãCãFãDäºç¹å ±å
â´â AFB=â ACB=45°=â AFD
â´AB²=AF²+BF²-2AF·BF·cos45°
AD²=AF²+DF²-2AF·DF·cos45°
BF²-â2·AF·BF=DF²-â2·AF·DF
BF²-DF²=(BF+DF)(BF-DF)=â2·AF(BF-DF)
â´BF+DF=â2·AF