В треугольнике ABC биссектриса -AD, причем AD=DC, угол C=20 см. Найти углы треугольников ABC иADC

Ответ:
1). AD = DC, значит, треугольник ADC - равнобедренный и http://tex.z-dn.net/?f=%5Cangle+DAC+%3D+%5Cangle+DCA+%3D+20%5E%7B%5Ccirc%7D" id="TexFormula1" onerror="texError(this)" title="\angle DAC = \angle DCA = 20^{\circ}" alt="\angle DAC = \angle DCA = 20^{\circ}" align="absmiddle" class="latex-formula">. 
http://tex.z-dn.net/?f=%5Cangle+ADC+%3D+180%5E%7B%5Ccirc%7D+-+%28%5Cangle+DAC+%2B+%5Cangle+DCA%29+%3D+180%5E%7B%5Ccirc%7D+-+2%2A20%5E%7B%5Ccirc%7D+%3D+140%5E%7B%5Ccirc%7D" id="TexFormula2" onerror="texError(this)" title="\angle ADC = 180^{\circ} - (\angle DAC + \angle DCA) = 180^{\circ} - 2*20^{\circ} = 140^{\circ}" alt="\angle ADC = 180^{\circ} - (\angle DAC + \angle DCA) = 180^{\circ} - 2*20^{\circ} = 140^{\circ}" align="absmiddle" class="latex-formula">.

2). AD - биссектриса, значит, http://tex.z-dn.net/?f=%5Cangle+BAD+%3D+%5Cangle+DAC+%3D+20%5E%7B%5Ccirc%7D" id="TexFormula3" onerror="texError(this)" title="\angle BAD = \angle DAC = 20^{\circ}" alt="\angle BAD = \angle DAC = 20^{\circ}" align="absmiddle" class="latex-formula"> и
http://tex.z-dn.net/?f=%5Cangle+A+%3D+2%2A+%5Cangle+DAC+%3D+2%2A20%5E%7B%5Ccirc%7D+%3D+40%5E%7B%5Ccirc%7D" id="TexFormula4" onerror="texError(this)" title="\angle A = 2* \angle DAC = 2*20^{\circ} = 40^{\circ}" alt="\angle A = 2* \angle DAC = 2*20^{\circ} = 40^{\circ}" align="absmiddle" class="latex-formula">.
http://tex.z-dn.net/?f=%5Cangle+B+%3D+180%5E%7B%5Ccirc%7D+-+%28%5Cangle+C+%2B+%5Cangle+A%29+%3D+180%5E%7B%5Ccirc%7D+-+%2820%5E%7B%5Ccirc%7D+%2B+40%5E%7B%5Ccirc%7D%29%3D+120%5E%7B%5Ccirc%7D" id="TexFormula5" onerror="texError(this)" title="\angle B = 180^{\circ} - (\angle C + \angle A) = 180^{\circ} - (20^{\circ} + 40^{\circ})= 120^{\circ}" alt="\angle B = 180^{\circ} - (\angle C + \angle A) = 180^{\circ} - (20^{\circ} + 40^{\circ})= 120^{\circ}" align="absmiddle" class="latex-formula">.

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