∫△AMC and△△△ CNB are isosceles right triangles,
∴DC=MD,EB=NE,
Let AC=xcm, then BC=(8-x)cm,
∴ The sum of the areas of two triangles is S= 12? x? 12x+ 12? (8-x)? 12(8 x)
= 14 x2+ 14(64+x2- 16x)
= 14 x2+ 16+ 14 x2-4x
= 12x2-4x+ 16
When AC=-? 42× 12=4,
That is, when BC=8-4=4cm, the areas of the two isosceles triangles are the smallest.