Toán hình lớp 7.

D

duc_2605

$\widehat{EAD} = 180^0 - (\widehat{AED} + \widehat{ADE})$
-----------------$= 180^0 - (\widehat{BAD} + \widehat{EAC})$
-----------------$= 180^0 - (\widehat{BAE} + 2\widehat{EAD} + \widehat{EAC})$
-----------------$= 90^o - \widehat{EAD}$
\Rightarrow $\widehat{EAD} = \dfrac{90^o}{2} = 45^0$
 
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