1、(C).
2、100.
3、6.
4、根据题意,得△AED ≌ △ACD,
AE = AC = 6,ED = CD,
∠AED = ∠C = 90°.由勾股定理,
得AB = 10.设ED = CD = x.在Rt△BDE中
DE² + EB² = DB²,
即x² + (10 - 6)² = (8 - x)² .
解得x = 3,即CD = 3 cm.
5、连接CE.
∵ ∠A = 90°,
∴ EC² = AC² + AE².
∵ DE是BC的垂直平分线,
∴ EC = EB.
∴ BE² = AC² + AE².