勾股定理在RTΔACD中,AC^2=AD^2+CD^2=64+CD^2,在RTΔBCD中,BC^2=BD^2+CD^2=4+CD^2,在RTΔABC中,AC^2+BC^2=AB^2=100,∴68+2CD^2=100,CD^2=16,CD=4。相似解法解:∵CD⊥AB,∠ACB=90°∴△ACD ∽△CBD∴CD ²=AD ×BD∵BD=2,AD=8∴CD²=16∴CD =4cm没有图,不知道我的理解对不亲,给个好评。两种解法