Đáp án:$ΔABC$ có $\widehat{ABC} = 36^o, \widehat{ABC} = \widehat{ACB} =72^o$
Giải thích các bước giải:
$+)$ Xét $ΔABC$ cân tại $A$
$⇒\widehat{BCA} = 90^o-\dfrac{\widehat{BAC}}{2}$
$+)$ Xét $ΔADC$ có $AD = DC$ $(gt)$
$⇒ΔADC$ cân tại $D$
$⇒ \left\{ \begin{array}{l}\widehat{ADC} = 180^o-2.\widehat{ACD}(1)\\\widehat{DAC} = \widehat{DCA}\end{array} \right.$
$+)$ Xét $ΔBDC$ có $DC = BC$ $(gt)$
$⇒ΔBDC$ cân tại $C$
$⇒\widehat{BDC} = 90^o - \dfrac{\widehat{BCD}}{2}$ $(2)$
Từ $(1)$ và $(2)$ suy ra :
$\widehat{ADC}+\widehat{BDC} = 180^o - 2.\widehat{ACD}+ 90^o -\dfrac{\widehat{BCD}}{2}$
$⇔180^o = 270^o - 2.\widehat{ACD}- \dfrac{\widehat{BCD}}{2}$
$⇔ 2.\widehat{ACD}+ \dfrac{\widehat{BCD}}{2} = 90^o$
$⇔ 4.\widehat{ACD}+ \widehat{BCD} = 180^o$
$⇔3.\widehat{ACD} + (\widehat{ACD}+ \widehat{BCD}) = 180^o$
$⇔3.\widehat{ADC} + \widehat{BCA} = 180^o$
$⇔3.\widehat{BAC} + 90^o-\dfrac{\widehat{BAC}}{2} = 180^o$
$⇔\dfrac{5.\widehat{BAC}}{2} = 90^o$
$⇔\widehat{BAC} = 36^o$
$⇒\widehat{ABC} = \widehat{ACB} = 180^o-2.\widehat{BAC} = 72^o$
Vậy $ΔABC$ có $\widehat{ABC} = 36^o, \widehat{ABC} = \widehat{ACB} =72^o$
