a)

$\,\,\,\,\,\,\widehat{BAD}=\widehat{CAE}=90{}^\circ $

$\to \widehat{BAD}+\widehat{BAC}\,\,=\,\,\widehat{CAE}+\widehat{BAC}$

$\to \widehat{DAC}=\widehat{BAE}$

Xét $\Delta ADC$ và $\Delta ABE$, ta có:

$\begin{cases}AD=AB\,\,\left(\,gt\,\right)\\AC=AE\,\,\left(\,gt\,\right)\\\widehat{DAC}=\widehat{BAE}\,\,\left(\,cmt\,\right)\end{cases}$

$\to \Delta ADC=\Delta ABE$

$\to DC=BE$  ( hai cạnh tương ứng )

b)

Trên tia $AM$ lấy điểm $K$ sao cho $MK=MA$

Xét $\Delta MDK$ và $\Delta MEA$, ta có:

$\begin{cases}MD=ME\,\,\left(\,gt\,\right)\\MK=MA\,\,\left(\,gt\,\right)\\\widehat{DMK}=\widehat{EMA}\,\,\left(\,\text{hai góc đối đỉnh}\,\right)\end{cases}$

$\to \Delta MDK=\Delta MEA$

$\to \widehat{MDK}=\widehat{MEA}$ ( hai góc tương ứng )

Mà 2 góc này nằm ở vị trí so le trong

Nên $DK\,\,||\,\,AE$

$\to \widehat{ADK}+\widehat{DAE}=180{}^\circ \,\,\,\,\,\left( 1 \right)$ ( hai góc trong cùng phía )

$\,\,\,\,\,\,\widehat{BAC}+\widehat{BAD}+\widehat{DAE}+\widehat{EAC}=360{}^\circ $

$\to \widehat{BAC}+\widehat{DAE}=360{}^\circ -\widehat{BAD}-\widehat{EAC}$

$\to \widehat{BAC}+\widehat{DAE}=360{}^\circ -90{}^\circ -90{}^\circ $

$\to \widehat{BAC}+\widehat{DAE}=180{}^\circ \,\,\,\,\,\left( 2 \right)$

Từ $\left( 1 \right)$ và $\left( 2 \right)$, suy ra $\widehat{ADK}=\widehat{BAC}$

$\,\,\,\,\,\,\Delta MDK=\Delta MEA\,\,\left( \,cmt\, \right)$

$\to DK=AE$ ( hai cạnh tương ứng )

$\to DK=AC$ ( Vì $AE=AC$ )

Xét $\Delta ADK$  và  $\Delta BAC$ có:

$\begin{cases}AD=BA\,\,\left(\,gt\,\right)\\DK=AC\,\,\left(\,cmt\,\right)\\\widehat{ADK}=\widehat{BAC}\,\,\left(\,cmt\,\right)\end{cases}$

$\to \Delta ADK=\Delta BAC$

$\to \widehat{DAK}=\widehat{ABC}$  ( hai góc tương ứng )

$\,\,\,\,\,\,\widehat{BAH}+\widehat{BAD}+\widehat{DAK}=180{}^\circ $  ( ba góc kề bù )

$\to \widehat{BAH}+\widehat{DAK}=180{}^\circ -\widehat{BAD}$

$\to \widehat{BAH}+\widehat{DAK}=180{}^\circ -90{}^\circ $

$\to \widehat{BAH}+\widehat{DAK}=90{}^\circ $

Mà $\widehat{DAK}=\widehat{ABC}\,\,\left( \,cmt\, \right)$

Nên $\widehat{BAH}+\widehat{ABC}=90{}^\circ $

$\to AH\bot BC$

$\to AM\bot BC$