《2025年全程检测单元测试卷九年级数学全一册湘教版》

18. (12分)如图7,在$□ ABCD$中,对角线$AC$与$BD$相交于点$O$,过点$B$作$BE\perp AB$交$AC$于点$E$.
(1)(6分)当$AC$平分$\angle DAB$时,求证:$AC\perp BD$.
(2)(6分)在(1)的条件下,若$OA=10$,$\cos \angle OBE=\frac{5}{7}$,求线段$AB$的长.

答案： 18．
(1)$\because$四边形$ABCD$是平行四边形，$\therefore AD // BC, OA=OC. \therefore \angle DAC=\angle ACB$. $\because AC$平分$\angle DAB, \therefore \angle DAC=\angle CAB. \therefore \angle ACB=\angle CAB. \therefore AB=BC$. 又$\because OA=OC, \therefore AC \perp BD$.
(2)$\because AC \perp BD, \therefore \angle AOB=90°. \therefore \angle OAB+\angle OBA=90°. \because BE \perp AB, \therefore \angle ABE=90°. \therefore \angle ABO+\angle OBE=90°. \therefore \angle OAB=\angle OBE. \therefore \cos \angle OAB=\cos \angle OBE=\frac{5}{7}$. $\because$在$\mathrm{Rt}\triangle ABO$中，$\cos \angle OAB=\frac{OA}{AB}. \therefore \frac{OA}{AB}=\frac{5}{7}$, 即$\frac{10}{AB}=\frac{5}{7}. \therefore AB=14$.