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| a Let the angle EDG be made equal to A, | a 23. 1. |
| and the fide DG b = DF c = AC; and let EG, | b 1. |
| and FG be joined. | c hyp. |
| 1. Caſe. Iſ EG fall above EF ; becauſe AE d= | d hyp. |
| DE, and ACe = DG, and the angle Ae = EDG, | e conjtr. |
| f therefore is BC = EG. But becauſe DFe = DG, | f 4. 1. |
| g, therefore is the angle DFG = DGF; h therefore | g 5. 1. |
| is the angle DFG = FFG, and by conſequence | h 9. ax. |
| the angle EFGb = EGF, k wherefore EG(BC) | k 19. 1. |
| = EF. |
| width="80%" | width="20" | and the fide DG b = DF c = AC; and let EG,
and FG be joined.
1. Caſe. Iſ EG fall above EF ; becauſe AE,d, =
DE, and ACe = DG, and the angle Ae = EDG
f therefore is BC = EG. But becauſe DFe = DG,
g, therefore is the angle DFG = DGF; h therefore
is the angle DFG = FFG, and by conſequence
the angle EFGb = EGF, k wherefore EG(BC)
= EF. || e conjitr. |