《2026年龙江王中王中考总复习数学》

19. 如图，在$\triangle ABC$中，$AD \bot BC$，$AE$是边$BC$上的中线，$AB = 1 0$，$AD = 6$，$\tan \angle ACB = 1$.
(1)求$BC$的长；
(2)求$\sin \angle DAE$的值.

答案： 19. 解：
(1) $\because AD \perp BC, \therefore \angle ADB = \angle ADC = 90°$.
$\therefore BD = \sqrt{AB^2 - AD^2} = 8$.
$\because \tan \angle ACB = \frac{AD}{CD} = 1, \therefore CD = AD = 6$.
$\therefore BC = BD + CD = 8 + 6 = 14$.
(2) $\because AE$ 是边 $BC$ 上的中线, $\therefore CE = \frac{1}{2}BC = 7$.
$\therefore DE = CE - CD = 7 - 6 = 1$.
$\therefore AE = \sqrt{AD^2 + DE^2} = \sqrt{37}$.
$\therefore \sin \angle DAE = \frac{DE}{AE} = \frac{\sqrt{37}}{37}$.

20. 在$\triangle ABC$中，$D$是边$AB$上任意一点，$E$是边$BC$的中点，过点$C$作$AB$的平行线，交$DE$的延长线于点$F$，连接$BF$，$CD$.
(1)如图①，求证：四边形$CDBF$是平行四边形；
(2)如图②，若$\angle FDB = 3 0 ^ { \circ } $，$\angle ABC = 4 5 ^ { \circ } $，$BC = 4 \sqrt { 2 }$，求$DF$的长.

答案：
20.
(1) 证明：$\because CF // DB, \therefore \angle 1 = \angle 2, \angle 3 = \angle 4$.
$\because E$ 为 $BC$ 的中点, $\therefore CE = BE$.
在 $\triangle CEF$ 和 $\triangle BED$ 中, $\begin{cases} \angle 1 = \angle 2, \\ \angle 3 = \angle 4, \\ CE = BE, \end{cases}$
$\therefore \triangle CEF \cong \triangle BED (AAS). \therefore EF = DE$.
又 $\because CE = BE, \therefore$ 四边形 $CDBF$ 是平行四边形.

(2) 解：$\because CB = 4\sqrt{2}, CE = BE$,
$\therefore BE = \frac{1}{2}BC = \frac{1}{2} × 4\sqrt{2} = 2\sqrt{2}$.
过 $E$ 作 $EH \perp DB$ 于 $H$.
则 $\angle EHD = \angle EHB = 90°. \therefore \angle HEB = 45° = \angle ABC$.
$\therefore EH = BH = \cos 45° · BE = \frac{\sqrt{2}}{2}BE = 2$.
$\because \angle FDB = 30°, \therefore DE = 2EH = 4. \therefore DF = 2DE = 8$.