解?$：(1)$?由題可知
?$∠BAC=180°-∠B-∠C= 180°-30°-50°=100°$?
∵?$∠BAE=∠CAE=\frac 12∠BAC$?
∴?$∠CAE=\frac 12×100°=50°$?
∴?$∠AEC=180°-50°-50°=80°$?
∴?$∠DEH=80°$?
?$∠EDH=90° -80°=10°$?
?$(2)∠C-∠B=2∠EDH$?
證明?$：∠C=180°-∠CAE-∠AEC$?
?$∠B=180°-∠BAD-∠AEB$?
?$∠C-∠B=180°-∠CAE-∠AEC-(180°-∠BAD-∠AEB)$?
?$=∠AEB-∠AEC=180°-∠AEC-∠AEC=180°-2∠AEC$?
?$=2(90°-∠AEC)=2(90°-∠DEH)=2∠EDH$?
