由A作垂线交BC于H．设∠BAE=y，设BH=AH=CH=1．则
EH=tan（45-y）=
| 1−tany |
| 1+tany |
HF=tany
EF=EH+HF=
| 1−tany |
| 1+tany |
BE=1-EH=
| 2tany |
| 1+tany |
CF=1-tany
令x=tany，则
EF=x+
| 1−x |
| 1+x |
BE=
| 2x |
| 1+x |
CF=1-x
CF2+BE2=（1-x）2+（
| 2x |
| 1+x |
| 1−x |
| 1+x |
故这三条线段可做成直角三角形．
由A作垂线交BC于H．| 1−tany |
| 1+tany |
| 1−tany |
| 1+tany |
| 2tany |
| 1+tany |
| 1−x |
| 1+x |
| 2x |
| 1+x |
| 2x |
| 1+x |
| 1−x |
| 1+x |