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(1)证明:作AG平分∠BAC,交BD于点G
∵∠BAC=90°,AE⊥BD,
∴∠DAE+∠ADB=∠ABE+∠ADB=90°,
∴∠ABG=∠CAF,
∵△ABC是等腰直角三角形,
∴AB=AC,∠C=∠BAG=45°,
∴
| ∠ABG=∠CAF | AB=AC | ∠C=∠BAG=45° |
| |
∴△BAG≌△CAF,(ASA)
∴AG=CF,
又∵AD=CD,∠GAD=∠C=45°,
∴△AGD≌△DFC,(SAS)
∴∠ADB=∠CDF;
(2)解:∠ADB=∠CMF.
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证明:作AG平分∠BAC,交BD于点G
∵∠BAC=90°,AE⊥BD,
∴∠DAE+∠ADB=∠ABE+∠ADB=90°,
∴∠ABG=∠CAF,
∵△ABC是等腰直角三角形,
∴AB=AC,∠C=∠BAG=45°,
∴
| ∠ABG=∠CAF | AB=AC | ∠C=∠BAG=45° |
| |
∴△BAG≌△CAF,(ASA)
∴AG=CF,
又∵AD=CM,∠GAD=∠C=45°,
∴△AGD≌△CFM,(SAS)
∴∠ADG=∠CMF;
即:∠ADB=∠CMF.