求完整过程,有好评😊
来源:学生作业帮 编辑:搜搜做题作业网作业帮 分类:历史作业 时间:2024/07/01 00:01:27
求完整过程,有好评😊
![](http://img.wesiedu.com/upload/5/d5/5d574687834148ae952a514383556b8f.jpg)
![](http://img.wesiedu.com/upload/5/d5/5d574687834148ae952a514383556b8f.jpg)
![求完整过程,有好评😊](/uploads/image/z/1767366-54-6.jpg?t=%E6%B1%82%E5%AE%8C%E6%95%B4%E8%BF%87%E7%A8%8B%2C%E6%9C%89%E5%A5%BD%E8%AF%84%26%23128522%3B)
问题在哪里呃.
再问:![](http://img.wesiedu.com/upload/f/9a/f9abc203ef28bc4eccc487d84d816428.jpg)
再答: 1.证明: ∵△ADC和 △ABC都是直角三角形 又∵M为AC的中点 ∴BM = DM = 1/2AC ∵DB⊥MN,MD=MD ∴△DMO ≌ △BMO ∴∠DMN = ∠BMN ∵DM‖BN ∴∠DMN = ∠BNM ∴∠BMN = ∠BNM ∵BO = BO ∴△BMO ≌ △BNO ∴BM = BN = DM ∴DM 平行且等于 BN ∴四边形DMBN为平行四边形 ∵MN⊥BD ∴平行四边形DMBN为菱形 2. ∵ AM = BM = DM ∴ ∠BAM = ∠MBA, ∠DAM = ∠MDA ∴ ∠BMC = 2∠BAM = 2*30° = 60° ∠DMC = 2∠DAM = 2*45° = 90° ∴ ∠DMB = ∠BMC+∠DMC = 90+60 = 150° ∴ ∠MDN = 180°-∠DMB = 180-150 = 30°
再问:
![](http://img.wesiedu.com/upload/f/9a/f9abc203ef28bc4eccc487d84d816428.jpg)
再答: 1.证明: ∵△ADC和 △ABC都是直角三角形 又∵M为AC的中点 ∴BM = DM = 1/2AC ∵DB⊥MN,MD=MD ∴△DMO ≌ △BMO ∴∠DMN = ∠BMN ∵DM‖BN ∴∠DMN = ∠BNM ∴∠BMN = ∠BNM ∵BO = BO ∴△BMO ≌ △BNO ∴BM = BN = DM ∴DM 平行且等于 BN ∴四边形DMBN为平行四边形 ∵MN⊥BD ∴平行四边形DMBN为菱形 2. ∵ AM = BM = DM ∴ ∠BAM = ∠MBA, ∠DAM = ∠MDA ∴ ∠BMC = 2∠BAM = 2*30° = 60° ∠DMC = 2∠DAM = 2*45° = 90° ∴ ∠DMB = ∠BMC+∠DMC = 90+60 = 150° ∴ ∠MDN = 180°-∠DMB = 180-150 = 30°