《2025年基础训练大象出版社八年级数学上册人教版》

17. (8分)如图,在$\triangle ABC$中,$D为边BC$的中点,过点$B作BE// AC交AD的延长线于点E$.
(1)求证:$\triangle BDE\cong\triangle CDA$;
(2)若$AD\perp BC$,求证:$BA = BE$.

答案：
(1) $\because$ D为边BC的中点，$\therefore BD = CD$.$\because BE// AC$,$\therefore \angle EBD = \angle C$, $\angle E = \angle CAD$.在$\triangle BDE$和$\triangle CDA$中，$\left\{\begin{array}{l} \angle EBD = \angle C\\ \angle E = \angle CAD\\ BD = CD\end{array}\right.$$\therefore \triangle BDE \cong \triangle CDA(AAS)$.
(2) $\because$ D为边BC的中点，$AD\perp BC$,$\therefore$ 直线AD为线段BC的垂直平分线.$\therefore BA = CA$.由
(1)知，$\triangle BDE \cong \triangle CDA$,$\therefore BE = CA$.$\therefore BA = BE$.

18. (8分)如图,已知在$\triangle ABC$中,$\angle BAC = 45^{\circ}$,$AD\perp BC于点D$,$BE\perp AC于点E$,$AD与BE相交于点F$.
(1)线段$AF与BC的数量关系是AF$=(填“$＞$”“$＜$”或“$=$”) $BC$.
(2)若$AF = 2BD$,求$\angle ABC$的度数.

(2) $\because AF = 2BD$, $AF = BC$,$\therefore BC = 2BD$.$\therefore$ D为BC的中点.$\because AD\perp BC$,$\therefore$ AD垂直平分BC;$\therefore AB = AC$;$\therefore \angle C = \angle ABC$.$\because \angle BAC = 45^{\circ}$,$\therefore \angle ABC = \frac{1}{2}(180^{\circ} - \angle BAC) = \frac{1}{2}×(180^{\circ} - 45^{\circ}) = 67.5^{\circ}$.

答案：
(1) =
(2) $\because AF = 2BD$, $AF = BC$,$\therefore BC = 2BD$.$\therefore$ D为BC的中点.$\because AD\perp BC$,$\therefore$ AD垂直平分BC;$\therefore AB = AC$;$\therefore \angle C = \angle ABC$.$\because \angle BAC = 45^{\circ}$,$\therefore \angle ABC = \frac{1}{2}(180^{\circ} - \angle BAC) = \frac{1}{2}×(180^{\circ} - 45^{\circ}) = 67.5^{\circ}$.