$解：作??AG⊥BC，??垂足為??G??$
$∵??AD=AC??$
$∴??∠ACB=∠FDC??$
$∵點(diǎn)??D??是??BC??的中點(diǎn)$
$∴??DB=DC??$
$∵??DE⊥BC??$
$∴??∠EDB=∠EDC=90°??$
$在??△BDE??和??△CDE??中$
$??\begin{cases}{DE=DE}\\{∠EDB=∠EDC}\\{DB=DC}\end{cases}??$
$∴??△BDE≌△CDE(\mathrm {SAS})??$
$∴??∠FCD=∠ABC??$
$∵??∠FDC=∠ACB??$
$∴??△FCD∽△ABC，??且相似比為??CD：????BC=1：????2??$
$∴??S_{△ABC}=4S_{△FCD}??$
$∵??S_{△FCD}=5??$
$∴??S_{△ABC}=\frac 12×BC×AG=20??$
$∵??BC=10??$
$∴??AG=4??$
$∵點(diǎn)??D??為??BC??的中點(diǎn)$
$∴??BD=CD=5??$
$∵??AD=AC，????AG⊥BC??$
$∴點(diǎn)??G??為??CD??的中點(diǎn)，??DG=\frac 12CD=\frac 52??$
$∴??BG=BD+DG=\frac {15}{2}??$
$∵??DE⊥BC??$
$∴??DE//AG??$
$∴??△BDE∽△BGA??$
$∴??\frac {BD}{BG}=\frac {DE}{AG}??$
$∵??BD=5，????BG=\frac {15}{2}，????AG=4??$
$∴??\frac 5{\frac {15}{2}}=\frac {DE}4??$
$∴??DE=\frac 83??$
