\(CD = BD + AB\)
证明：在\(CD\)上截取\(DH = DB\)，连接\(AH\)，
\(\because AD \bot BC\)，
\(\therefore AB = AH\)，
\(\therefore \angle AHB = \angle B\)，
\(\because \angle B = 2\angle C\)，
\(\therefore \angle AHB = \angle C\)，
\(\because \angle AHB = \angle C + \angle HAC\)，
\(\therefore \angle HAC = 2\angle C\)，
\(\therefore AH = CH\)，
\(\therefore AB = CH\)，
\(\therefore AB + BD = CH + DH = CD\)．
