12.(10分)如图,在△ABC中,AD平分∠BAC,过点B作AD的垂线,垂足为D,作DE//AC,交AB于点E,连接CD,CD//AB.
(1)求证:△BDE是等腰三角形;
(2)求证:CD=BE.
答案：12.证明:(1)如答图，
$\because DE// AC,\therefore ∠ 1=∠ 4.$
$\because AD$平分$∠ BAC,\therefore ∠ 1=∠ 2,\therefore ∠ 2=∠ 4.$
$\because AD⊥ BD,\therefore ∠ 2+∠ ABD=90°,∠ 5+∠ 4=90°,$
$\therefore ∠ 5=∠ ABD,\therefore DE=BE,$
$\therefore △ BDE$是等腰三角形.
(2)由(1)知,$∠ 1=∠ 2,∠ 2=∠ 4.$
$\because CD// AB,\therefore ∠ 2=∠ 3,\therefore ∠ 3=∠ 4.$
又$\because AD=AD,\therefore △ ACD≌ △ AED(\mathrm{ASA}),$
$\therefore CD=DE,$由(1)知$DE=BE,\therefore CD=BE.$
