-pà con nào giỏi hình giup tớ với

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truongtrang12

Ta có : [TEX]\hat{BAC} + \hat{ABC} + \hat{ACB} = 180°[/TEX]

-> [TEX]\hat{ABC} + \hat{ACB} = 180° - 130° =50°[/TEX]

Vì AB = BD -> tam giác ABD cân tại B

-> [TEX]\hat{ADB} = \frac{180° - \hat{B}}{2} = \frac{180° - ( 50° - \hat{C})}{2}= \frac{130°+\hat{C}}{2}[/TEX]

Vì AC = CE -> tam giác ACE cân tại C

->[TEX] \hat{EAC} = \frac{180° - \hat{C}}{2} = \frac{180° - ( 50° - \hat{B})}{2} = \frac{130° +\hat{B}}{2}[/TEX]

[TEX]\hat{EAC} + \hat{ADB} = \frac{130° +\hat{C} }{2} + \frac{130° +\hat{B}}{2} = \frac{130°+130°+50°}{2} = 155°[/TEX]

ta có [TEX]\hat{EAC} + \hat{ADB} =( \hat{BAE} + \hat{CAD} + \hat{EAD} ) + \hat{EAD} = 155°[/TEX]

[TEX]=130° +\hat{EAD} = 155°[/TEX]

->[TEX]\hat{EAD} = 155° -130°[/TEX]

->[TEX]\hat{EAD} = 25°[/TEX]
 
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