證明：?$(1)$?連接?$OD. $?
?$∵ OA=OD，$?
?$∴ ∠OAD=∠ODA.$?
?$∵ AD$?平分?$∠BAC，$?
?$∴ ∠OAD=∠BAD，$?
?$∴ ∠ODA=∠BAD，$?
?$∴ OD//AB，$?
?$∴ ∠ODC=∠B=90°，$?
?$∴ OD⊥BC.$?
?$∵ OD$?是?$⊙O$?的半徑，
?$∴ BC$?是?$⊙O$?的切線.
?$(2)$?連接?$OF、$??$DE.$?
?$∵ ∠B=90°，$??$∠ADB=60°，$?
?$∴ ∠BAD=30°，$?
?$∴ AD=2BD=10. $?
?$∵ AE$?是?$⊙O$?的直徑，
?$∴ ∠ADE=90°. $?
?$∵ AD$?平分?$∠BAC，$?
?$∴ ∠DAE=∠BAD=30°，$?
?$∴ DE= \frac {1}{2}\ \mathrm {AE}. $?
∵ 在?$Rt∠ADE$?中，?$DE2+AD2=AE2，$?
?$∴ ( \frac {1}{2}\ \mathrm {AE})2+102=AE2，$?
解得?$AE =\frac {20\sqrt{3}}{3} ($?負(fù)值舍去)，
?$∴ OA= \frac {1}{2}\ \mathrm {AE}= \frac {10\sqrt{3}}{3} . $?
?$∵ AD$?平分?$∠BAC，$?
?$∴ ∠BAC=2∠BAD=60°. $?
?$∵ OA=OF，$?
?$∴ △AOF$?是等邊三角形，
?$∴ ∠AOF=60°. $?
?$∵ OD//AB，$?
?$∴ S_{△ADF}=S_{△AOF}，$?
?$∴ S_{涂色}=S_{扇形}OAF= \frac {60π×(\frac {10\sqrt{3}}{3})2}{360}=\frac {50π}{9}.$?
