$證明:(1)∵AE⊥AD,EH⊥AC$
$∴ ∠EHA=∠EAD=∠ACB=90°$
$∴∠DAC+∠ADC=90°$
$∠DAC+∠EAH=90°$
$∴∠ADC=∠EAH$
$在△ADC和△EAH中$
$\begin{cases}{ ∠ACD=∠EHA} \\ { ∠ADC=∠EAH} \\{ AD=EA} \end{cases}$
$∴△ADC≌△EAH(AAS),∴EH=AC$
$(2)如圖,過點(diǎn)E作EN⊥AM,交AM的延長線于點(diǎn)N$
$∵AE⊥ AD,EN⊥AM,∴∠EAD=∠ANE=∠ACB=90°$
$∴∠ACD=180°?∠ACB=180°?90°=90°$
$∴∠ACD=∠ENA,∠DAC+∠ADC=90°$
$∵∠DAC+∠EAN=180°?∠EAD=180°?90°=90°$
$∴∠ADC=∠EAN$
$在△ADC 和△EAN 中$
$\begin{cases}{∠ACD=∠ENA\ } \\ {∠ADC=∠EAN\ } \\{ AD=EA} \end{cases}$
$∴ △ADC≌△EAN(AAS),∴EN=AC$
$∵AC=BC,∴EN=BC$
$∵S_{△ABM}=\frac{1}{2}×AM×BC,S_{△AME}=\frac{1}{2}×AM×EN$
$∴S_{△ABM}=S_{△AME}$
