∵∠OAD+∠ODA=90°,∠ADB+∠ABD=90°,∴∠DAO=∠DBA,
又∵∠AOD=∠DAB=90°,∴△ADO∽△BAD
∴AD/DO=BD/AD,变形得AD^2=DO*BD,
同理可得AB^2=BO*BD,
结论等式右边变为(DO*BD)/(BO*BD)即DO/BO,
∵AB∥CD,∴△ABO∽△CDO,∴CO/AO=DO/BO,得证.
∵∠OAD+∠ODA=90°,∠ADB+∠ABD=90°,∴∠DAO=∠DBA,
又∵∠AOD=∠DAB=90°,∴△ADO∽△BAD
∴AD/DO=BD/AD,变形得AD^2=DO*BD,
同理可得AB^2=BO*BD,
结论等式右边变为(DO*BD)/(BO*BD)即DO/BO,
∵AB∥CD,∴△ABO∽△CDO,∴CO/AO=DO/BO,得证.