AB⊥BD于点B,ED⊥BD于点D,AE交BD于点C,且BC=DC.求证AB=ED
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AB⊥BD于点B,ED⊥BD于点D,AE交BD于点C,且BC=DC.求证AB=EDAB⊥BD于点B,ED⊥BD于点D,AE交BD于点C,且BC=DC.求证AB=EDAB⊥BD于点B,ED⊥BD于点D,
AB⊥BD于点B,ED⊥BD于点D,AE交BD于点C,且BC=DC.求证AB=ED
AB⊥BD于点B,ED⊥BD于点D,AE交BD于点C,且BC=DC.求证AB=ED
AB⊥BD于点B,ED⊥BD于点D,AE交BD于点C,且BC=DC.求证AB=ED
证明：在△ABC与△EDC中
∠ABC=∠EDC=90度,BC=DC,∠ACB=∠ECD
所以△ABC≌△EDC
所以AB=ED
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证明：∵AB⊥BD，ED⊥BD
∴∠ABC=∠CDE=90°
在Rt△ABC和Rt△EDC中 ：
∠ABC=∠CDE、BC=CD 、∠ACB= ∠DCE
∴Rt△ABC≌Rt△EDC（ASA）
∴AB=ED
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