过点$D$作$DH\bot AB$于点$H$.

$\because AD$平分$\angle CAB$，$DC\bot AC$，$DH\bot AB$，
$\therefore \angle CAD=\angle DAH$，$\angle C=\angle AHD=90^{\circ}$，
在$\triangle ACD$和$\triangle AHD$中，
$\left\{\begin{array}{l}{∠C=∠AHD}\\{∠CAD=∠HAD}\\{AD=AD}\end{array}\right.$，
$\therefore \triangle ACD\equiv \triangle AHD\left(AAS\right)$，
$\therefore CD=DH$，$AC=AH$，
设$CD=DH=a$，
$\because CA=CB$，
$\therefore \angle ABC=\angle BDH=45^{\circ}$，
$\therefore DH=HB=a$，
$\therefore DB=\sqrt{2}a$，
$\therefore AC=BC=(1+\sqrt{2})a$，
$\therefore AD=\sqrt{A{C}^{2}+C{D}^{2}}=\sqrt{[(1+\sqrt{2})a]^{2}+{a}^{2}}=\sqrt{4+2\sqrt{2}}a$，$AB=\sqrt{2}AC=(2+\sqrt{2})a$，
$\because S_{\triangle ADB}=\frac{1}{2}\cdot AB\cdot DH=\frac{1}{2}AD\cdot BE$，
$\therefore \frac{1}{2}\times (2+\sqrt{2})a\times a=\frac{1}{2}\cdot \sqrt{4+2\sqrt{2}}\cdot a\cdot \frac{25}{4}$，
$\because a\neq 0$，
$\therefore a=\frac{25\sqrt{4+2\sqrt{2}}}{4(2+\sqrt{2})}$，
$\therefore AD=\frac{25×(2+2\sqrt{2})}{4(2+\sqrt{2})}=\frac{25\sqrt{2}}{4}$.
故答案为：$\frac{25\sqrt{2}}{4}$

过点$D$作$DH\bot AB$于点$H$.

$\because AD$平分$\angle CAB$，$DC\bot AC$，$DH\bot AB$，
$\therefore \angle CAD=\angle DAH$，$\angle C=\angle AHD=90^{\circ}$，
在$\triangle ACD$和$\triangle AHD$中，
$\left\{\begin{array}{l}{∠C=∠AHD}\\{∠CAD=∠HAD}\\{AD=AD}\end{array}\right.$，
$\therefore \triangle ACD\equiv \triangle AHD\left(AAS\right)$，
$\therefore CD=DH$，$AC=AH$，
设$CD=DH=a$，
$\because CA=CB$，
$\therefore \angle ABC=\angle BDH=45^{\circ}$，
$\therefore DH=HB=a$，
$\therefore DB=\sqrt{2}a$，
$\therefore AC=BC=(1+\sqrt{2})a$，
$\therefore AD=\sqrt{A{C}^{2}+C{D}^{2}}=\sqrt{[(1+\sqrt{2})a]^{2}+{a}^{2}}=\sqrt{4+2\sqrt{2}}a$，$AB=\sqrt{2}AC=(2+\sqrt{2})a$，
$\because S_{\triangle ADB}=\frac{1}{2}\cdot AB\cdot DH=\frac{1}{2}AD\cdot BE$，
$\therefore \frac{1}{2}\times (2+\sqrt{2})a\times a=\frac{1}{2}\cdot \sqrt{4+2\sqrt{2}}\cdot a\cdot \frac{25}{4}$，
$\because a\neq 0$，
$\therefore a=\frac{25\sqrt{4+2\sqrt{2}}}{4(2+\sqrt{2})}$，
$\therefore AD=\frac{25×(2+2\sqrt{2})}{4(2+\sqrt{2})}=\frac{25\sqrt{2}}{4}$.
故答案为：$\frac{25\sqrt{2}}{4}$