Question

在梯形ABCD中,AB∥CD,AD＝CD＝BC,∠ABC＝72°,AC,BD相较于点O.求证AB＝BD,点O是线段BD的

Answer 1

证明:(1)AB∥CD,AD=BC,则梯形ABCD为等腰梯形,故∠DAB=∠ABC=72°;

BC=CD,则∠CBD=∠CDB;

AB∥CD,则∠ABD=∠CDB.

故∠CBD=∠ABD=36°,∠ADB=180°-∠ABD-∠DAB=72°.

即∠ADB=∠DAB,得AB=BD.

(2)同理可求:∠DAC=36°=∠ABD;又∠ADO=∠ADB.

则⊿DAO∽⊿DBA,AD/BD=OD/AD,AD^2=BD*OD;

又∠AOD=180°-∠DAC-∠ADB=72°=∠ADO,AD=AO;

而∠DBA=∠CAB=36°,故BO=AO=AD.

所以,BO^2=DB*DO,故点O为线段BD的黄金分割点.