证明：以AD为对称轴作△ABD的轴对称△AED
          即∠B=∠AED=2∠C,BD=DE
          又在△ACE中,∠AED=∠C+∠CAE
         即∠C=∠CAE,△AEC为等腰三角形
     故AE=CE=AB
         即CD=DE+CE＝AB+BD
         证明完毕