数学题初一 急!

（1）∵角AED=角EDC+角ECD=15+角C
又角AED=角ADE
∴∠ADE=15+∠C
又∠ADC=ADE+EDC=B=+BAD
∴B+BAD=15+C+15=30+C
BAD=30+C-B=30
(2)2CDE=DAB
∵∠ACD=∠AED+CDE
所以∠abc=∠AED+∠cde
又角abc=角dab+角ADB
所以角dab+角ADB=角AED+角cde
角dab=AED+cde-ADB
又AED=ade
所以dab=ade+cde-ADB
∠DAB=2∠CDE