Đáp án:
$\begin{cases}\widehat{A} = 108^o\\\widehat{B} = \widehat{C} = 36^o\end{cases}$
Giải thích các bước giải:
Ta có: $ΔABC$ cân tại $A$ có:
$AH\perp BC$
$\Rightarrow \begin{cases}BH = HC\\\widehat{HAB} = \widehat{HAC} = \dfrac{1}{2}\widehat{BAC}\end{cases}$
Xét $ΔBDC$ có:
$HE//BD \, (gt)$
$BH = HC$
$\Rightarrow HE$ là đường trung bình
$\Rightarrow HE = \dfrac{1}{2}BD$
$\Rightarrow HE = HA$
$\Rightarrow ΔHAE$ cân tại $H$
$\Rightarrow \widehat{HAE} = \widehat{HEA}$
Ta được: $\widehat{HEA} + \widehat{HEA} + \widehat{AHE} = 180^o$
$\Leftrightarrow 2\widehat{HAE} + \widehat{AHE} = 180^o$
$\Leftrightarrow \widehat{BAC} + \widehat{AHE} = 180^o$ $(1)$
Mặt khác:
$\widehat{AHE} + \widehat{CHE} = \widehat{AHC} = 90^o$
$\Leftrightarrow \widehat{AHE} + \widehat{DBC} = 90^o$
$\Leftrightarrow \widehat{AHE} + \widehat{ABD} = 90^o$
$\Leftrightarrow \widehat{ABD} = 90^o - \widehat{AHE}$
Ta được:
$\widehat{ABD} + \widehat{ADB} + \widehat{BAD} = 180^o$
$\Leftrightarrow 90^o - \widehat{AHE} + \widehat{AEH} + \widehat{BAC} = 180^o$
$\Leftrightarrow -\widehat{AHE} + \widehat{HAE} + \widehat{BAC} = 90^o$
$\Leftrightarrow \dfrac{3}{2}\widehat{BAC} - \widehat{AHE} = 90^o$ $(2)$
$(1)(2)\Rightarrow \dfrac{5}{2}\widehat{BAC} = 270^o$
$\Rightarrow \widehat{BAC} = 108^o$
$\Rightarrow \widehat{ABC} = \widehat{ACB} = \dfrac{180^o - \widehat{BAC}}{2} = \dfrac{180^o - 108^o}{2} = 36^o$
