$ (1)$证明：∵$\triangle ABD$和$\triangle ACE$都为等腰直角三角形，
$∠BAD = ∠CAE = 90°$
∴$AD = AB，$$AE = AC，$
$∠DAB + ∠BAC = ∠EAC + ∠BAC，$即$∠DAC = ∠BAE$
$ $在$\triangle DAC$和$\triangle BAE$中
$\begin {cases}AD = AB\\∠DAC = ∠BAE\\AC = AE\end {cases}$
∴$\triangle DAC≌\triangle BAE(S AS)$
$ (2)$解：$\triangle ABC$与$\triangle ADE$的面积相等，理由如下：
如图，过点$E$作$EN\perp AD，$交$DA$的延长线于点$N，$
过点$C$作$CM\perp AB，$垂足为$M，$则$∠AMC = ∠ANE = 90°$
∵$\triangle ABD$和$\triangle ACE$都是等腰直角三角形，
$∠BAD = ∠CAE = 90°，$∴$AB = AD，$$AC = AE$
∵$∠BAD + ∠CAE + ∠BAC + ∠DAE = 360°$
∴$∠BAC + ∠DAE = 360°-∠BAD - ∠CAE = 180°$
∵$∠DAE+∠EAN = 180°，$∴$∠BAC = ∠EAN$
$ $在$\triangle ACM$和$\triangle AEN$中
$\begin {cases}∠AMC = ∠ANE\\∠CAM = ∠EAN\\AC = AE\end {cases}$
∴$\triangle ACM≌\triangle AEN(\mathrm {AAS})$
∴$CM = EN$
∵$S_{\triangle ABC}=\frac 12\ \mathrm {A}B·CM，$$S_{\triangle ADE}=\frac 12\ \mathrm {A}D·EN$
∴$S_{\triangle ABC}=S_{\triangle ADE}$
