八年级勾股定理证明题在RtΔABC中,∠ACB＝90°BC＝3,AC＝4,AB的垂直平分线DE交BC的延长线于点E,则C

连接AE,∵DE垂直平分AB,∴AE=BE,设CE=x,EB=3+x,在直角ΔACE中,x²+16=AE²,∴x²+16=(3+x)²
X=7/6,即CE=7/6