$?(1)?證明：∵?DO⊥AB ?$
$∴?∠DOB=∠DOA=90°?$
$∴?∠DOB=∠ACB?$
$∵?∠B=∠B?$
$∴?△DOB∽△ACB?$
$?(2)?解：∵?∠ACB=90° ?$
$∴?AB=\sqrt {AC^2+BC^2}=\sqrt {6^2+8^2}=10?$
$∵?AD?平分?∠CAB，??DC⊥AC，??DO⊥AB ?$
$∴?DC=DO?$
$在?Rt△ACD?和?Rt△AOD?中$
$?\begin{cases}AD=AD\\DC=DO\end{cases}?$
$∴?Rt△ACD≌Rt△AOD(\mathrm {HL})?$
$?AC=AO=6?$
$設(shè)?BD=x，?則?DC=DO=8-x，??OB=AB-AO=4?$
$在?Rt△BOD?中，由勾股定理得，?DO^2+OB^2=BD^2，?即?(8-x)^2+4^2=x^2?$
$解得?x=5?$
$∴?BD?的長(zhǎng)為?5?$
$?(3)?∵點(diǎn)?B'?與點(diǎn)?B?關(guān)于直線?DO?對(duì)稱$
$∴?∠B=∠OB'D，??BO=B'O，??BD=B'D?$
$∵?∠B?是銳角 $
$∴?∠OB'D?也為銳角$
$∴?∠AB'D?為鈍角$
$∴當(dāng)?△AB'D?為等腰三角形時(shí)，?AB'=DB'?$
$∵?△DOB∽△ACB?$
$∴?\frac {OB}{BD}=\frac {BC}{AB}=\frac 8{10}=\frac 45?$
$設(shè)?BD=5x，?則?AB'=DB'=5x，??BO=B'O=4x?$
$∵?AB'+B'O+BO=AB?$
$∴?5x+4x+4x=10?$
$解得?x=\frac {10}{13}?$
$∴?BD=\frac {50}{13}?$
