Tính độ dài đoạn thẳng AE và chứng minh góc EDH= góc BCH

Ta có:
$\frac{AB}{AC} = \frac{BD}{CE} = \frac{AD}{AE}$
$\frac{4}{5} = \frac{BD}{CE} = \frac{2}{AE}$
$\Rightarrow BD = \frac{4}{5}CE$ và $AE = \frac{5}{2}$
Ta có:
$\frac{BD}{CE} = \frac{4}{5} = \frac{2}{AE} = \frac{2}{\frac{5}{2}} = \frac{4}{5}$
$\Rightarrow BD = 2, CE = 2.5$
Do đó, $DE = BD + CE = 2 + 2.5 = 4.5$
Gọi $x = \widehat{EDH} = \widehat{BCH}$
Ta có:
$\widehat{AED} = 180^{\circ} - \widehat{ADE} - \widehat{EDA} = 180^{\circ} - \widehat{ADE} - (180^{\circ} - \widehat{BDE}) = \widehat{ADE} - \widehat{BDE}$
$\widehat{AEC} = 180^{\circ} - \widehat{ACE} - \widehat{ECA} = 180^{\circ} - \widehat{ACE} - (180^{\circ} - \widehat{CED}) = \widehat{ACE} - \widehat{CED}$
Vậy ta có:
$\frac{\widehat{AED}}{\widehat{AEC}} = \frac{\widehat{ADE} - \widehat{BDE}}{\widehat{ACE} - \widehat{CED}} = \frac{180^{\circ} - \widehat{ADE} - \widehat{EDA}}{180^{\circ} - \widehat{ACE} - \widehat{ECA}} = \frac{180^{\circ} - \widehat{ADE} - (180^{\circ} - \widehat{BDE})}{180^{\circ} - \widehat{ACE} - (180^{\circ} - \widehat{CED})} = \frac{\widehat{BDE}}{\widehat{CED}} = \frac{BD}{CE} = \frac{4}{5}$
$\Rightarrow \widehat{AED} = \frac{4}{5} \widehat{AEC}$
$\Rightarrow x = \frac{4}{5} (180^{\circ} - x)$
$\Rightarrow 5x = 4(180^{\circ} - x)$
$\Rightarrow 5x = 720^{\circ} - 4x$
$\Rightarrow 9x = 720^{\circ}$
$\Rightarrow x = 80^{\circ}$
Vậy ta có: $\widehat{EDH} = \widehat{BCH} = 80^{\circ}$