由题设知:AC=AE,DE=CD,∠AED=∠BED=90°.
由勾股定理:
AB^2=AC^2+BC^2.
AB^2=6^2+8^2.
=100.
∵ CD=DE.
在Rt△BED中,
BD^2=DE^2+BE^2.
(BC-CD)^2=DE^2+(AB-AE).
(8-DE)=DE^2+(10-6)^2.
64-16DE+DE^2=DE^2+16.
16DE=64-16.
=48..
∴DE=3.
由题设知:AC=AE,DE=CD,∠AED=∠BED=90°.
由勾股定理:
AB^2=AC^2+BC^2.
AB^2=6^2+8^2.
=100.
∵ CD=DE.
在Rt△BED中,
BD^2=DE^2+BE^2.
(BC-CD)^2=DE^2+(AB-AE).
(8-DE)=DE^2+(10-6)^2.
64-16DE+DE^2=DE^2+16.
16DE=64-16.
=48..
∴DE=3.