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	2	thêm 240 ký tự trong đáp án		Sửa 05-11-13 09:04 PM ♥♥♥ Panda Sơkiu Panda Mập ♥♥♥ 1
$\widehat{CED} = 180^{0} - \widehat{ECD} + \widehat{ECD}$ = $180^{0} - \frac{1}{2} ( \widehat{BCD} + \widehat{DBC})$ = $180^{a} - \frac{1}{2}(360^{0} - \widehat{A} - \widehat{B} )$ = $180^{0} - \frac{1}{2} ( 360^{0} - 200^{0})$ = $100^{0}$ hai đường phân giác trong và ngoài của tam giác sẽ vuông góc với nhau=> $\widehat{ECH} = \widehat{EDH} = 90^{0}$ ta có $\widehat{ECH} + \widehat{EDH} + \widehat{CED} + \widehat{CHD} = 360^{0}$ từ đây tính được góc còn lại $\widehat{CED} = 180^{0} - \widehat{ECD} + \widehat{ECD}$ = $180^{0} - \frac{1}{2} ( \widehat{BCD} + \widehat{DBC})$ = $180^{a} - \frac{1}{2}(360^{0} - \widehat{A} - \widehat{B} )$ = $180^{0} - \frac{1}{2} ( 360^{0} - 200^{0})$ = $100^{0}$ $\widehat{CED} = 180^{0} - \widehat{ECD} + \widehat{ECD}$ = $180^{0} - \frac{1}{2} ( \widehat{BCD} + \widehat{DBC})$ = $180^{a} - \frac{1}{2}(360^{0} - \widehat{A} - \widehat{B} )$ = $180^{0} - \frac{1}{2} ( 360^{0} - 200^{0})$ = $100^{0}$ hai đường phân giác trong và ngoài của tam giác sẽ vuông góc với nhau=> $\widehat{ECH} = \widehat{EDH} = 90^{0}$ ta có $\widehat{ECH} + \widehat{EDH} + \widehat{CED} + \widehat{CHD} = 360^{0}$ từ đây tính được góc còn lại

	1	tạo mới		Trả lời 05-11-13 08:57 PM ♥♥♥ Panda Sơkiu Panda Mập ♥♥♥ 1
$\widehat{CED} = 180^{0} - \widehat{ECD} + \widehat{ECD}$ = $180^{0} - \frac{1}{2} ( \widehat{BCD} + \widehat{DBC})$ = $180^{a} - \frac{1}{2}(360^{0} - \widehat{A} - \widehat{B} )$ = $180^{0} - \frac{1}{2} ( 360^{0} - 200^{0})$ = $100^{0}$

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