Cho tam giác AEF đồng dạng tam giác ABC (hình vẽ) nếu AE = 3cm; EB = 2cm; AF = 4cm, thì FC
AB = AE + EB = 2 + 3 = 5(cm)
Vì ∆AEF ∽ ∆ABC nên \(\frac{{AE}}{{AB}} = \frac{{AF}}{{AC}}\)
Suy ra: AC = \(\frac{{AB.AF}}{{AE}} = \frac{{20}}{3}\left( {cm} \right)\)
FC = AC – AF = \(\frac{{20}}{3} - 4 = \frac{8}{3}\left( {cm} \right)\).