《2025年新课标教材同步导练八年级数学下册北师大版C版》

14. 已知：如图，BE平分∠ABC，CE平分∠ACD交BE于E.
(1)求证：AE平分∠FAC；
(2)若∠ABC = 64°，求∠AEC的度数.

答案：
(1) 过点$E$分别作$EM\perp AB$，$EN\perp AC$，$EG\perp BC$，垂足分别为$M$、$N$、$G$.
$\because BE$平分$\angle ABC$，$CE$平分$\angle ACD$，
$\therefore EM = EG$，$EN = EG$，$\therefore EM = EN$，
$\therefore$点$E$在$\angle FAC$的平分线上，
$\therefore AE$平分$\angle FAC$.
(2)$\angle AEC = 90^{\circ}-\frac{1}{2}\angle ABC = 58^{\circ}$.

15. (2021·黔东南)在四边形ABCD中，对角线AC平分∠BAD.
【探究发现】
(1)如图①，若∠BAD = 120°，∠ABC = ∠ADC = 90°. 求证：AD + AB = AC；
【拓展迁移】
(2)如图②，若∠BAD = 120°，∠ABC + ∠ADC = 180°.
①猜想AB、AD、AC三条线段的数量关系，并说明理由；
②若AC = 10，求四边形ABCD的面积.

答案：
(1) 证明：$\because AC$平分$\angle BAD$，$\angle BAD = 120^{\circ}$，
$\therefore \angle DAC=\angle BAC = 60^{\circ}$
$\because \angle ADC=\angle ABC = 90^{\circ}$
$\therefore \angle ACD=\angle ACB = 30^{\circ}$，
$\therefore AD=\frac{1}{2}AC$，$AB=\frac{1}{2}AC$.
$\therefore AD + AB = AC$.
(2)①$AD + AB = AC$，
理由：过点$C$分别作$CE\perp AD$交$AD$延长线于$E$，$CF\perp AB$于$F$. $\because AC$平分$\angle BAD$，$CE\perp AD$于$E$，$CF\perp AB$，$\therefore CF = CE$
$\because \angle ABC+\angle ADC = 180^{\circ}$，$\angle EDC+\angle ADC = 180^{\circ}$，
$\therefore \angle ABC=\angle EDC$，在$\triangle CED$和$\triangle CFB$中，
$\begin{cases}\angle CDE=\angle CBF\\\angle E=\angle CFB\\CE = CF\end{cases}$
$\therefore \triangle CED\cong\triangle CFB(AAS)$，$\therefore FB = DE$，
$\therefore AD + AB = AD + FB + AF = AD + DE + AF = AE + AF$，
在四边形$AFCE$中，由
(1)题知：$AE + AF = AC$，$\therefore AD + AB = AC$.
②在$Rt\triangle ACE$中，$\because AC$平分$\angle BAD$，$\angle BAD = 120^{\circ}$，$\therefore \angle DAC=\angle BAC = 60^{\circ}$，
$\therefore \angle ACE = 30^{\circ}$，$\therefore AE=\frac{1}{2}AC = 5$，
$\therefore CE=\sqrt{AC^{2}-AE^{2}} = 5\sqrt{3}$，
$\because CF = CE$，$AD + AB = AC$，
$S_{四边形ABCD}=\frac{1}{2}AD\cdot CE+\frac{1}{2}AB\cdot CF=\frac{1}{2}(AD + AB)\cdot CE=\frac{1}{2}AC\cdot CE=\frac{1}{2}\times10\times5\sqrt{3}=25\sqrt{3}$.