解：設(shè)腰長?$AB=AC=x，$?底邊長?$BC=y$?
由題意得：周長?$2x + y = 16，$?故?$y = 16 - 2x$?
∵?$AD$?是底邊?$BC$?上的高，∴?$BD = DC = \frac 12y = 8 - x$?
在?$Rt?ABD$?中，?$AB^2 = AD^2 + BD^2$?
即?$x^2 = 4^2 + (8 - x)^2，$?解得?$x = 5$?
則?$y = 16 - 2×5 = 6$?
∴?$?ABC$?各邊長為?$5，$??$5，$??$6$?
