《2025年轻松暑假复习加预习中国海洋大学出版社七年级数学鲁教版54制》

8.(2022·扬州)将一副直角三角板如图放置,若$EF// BC$,则$∠BND= $$105^{\circ}$$^{\circ }$.

答案： $105^{\circ}$

9.(浙江中考)如图所示,$AB// CD$,EF分别与AB,CD交于点B,F.若$∠E= $$30^{\circ },∠EFC= 130^{\circ }$,则$∠A= $$20^{\circ}$.

答案： $20^{\circ}$

10.如图所示,在$△ABC$中,AD是BC边上的高,AE,BF分别是$∠BAC$和$∠ABC$的角平分线,它们相交于点O,$∠AOB = 125^{\circ }$,求$∠CAD$的度数.
解:$\because \angle AOB = 125^{\circ}$，
　　$\therefore \angle OAB + \angle OBA =$$55^{\circ}$。
　　$\because AE$，$BF$分别是$\angle BAC$和$\angle ABC$的角平分线，它们相交于点$O$
　　$\therefore \angle BAC + \angle ABC = 2(\angle OAB + \angle OBA) =$$110^{\circ}$。$\therefore \angle C =$$70^{\circ}$。
　　$\because AD$是$BC$边上的高，
　　$\therefore \angle ADC =$$90^{\circ}$。
　　$\therefore \angle CAD =$$20^{\circ}$。
　即$\angle CAD$的度数是$20^{\circ}$。

答案： 解:$\because \angle AOB = 125^{\circ}$，
　　$\therefore \angle OAB + \angle OBA = 55^{\circ}$。
　　$\because AE$，$BF$分别是$\angle BAC$和$\angle ABC$的角平分线，它们相交于点$O$
　　$\therefore \angle BAC + \angle ABC = 2(\angle OAB + \angle OBA) = 110^{\circ}$。$\therefore \angle C = 70^{\circ}$。
　　$\because AD$是$BC$边上的高，
　　$\therefore \angle ADC = 90^{\circ}$。
　　$\therefore \angle CAD = 20^{\circ}$。
　即$\angle CAD$的度数是$20^{\circ}$。

11.已知$△ABC$中,$∠B= ∠C$,D为边BC上一点(不与B,C重合),点E为边AC上一点,$∠ADE= $$∠AED,∠BAC= 44^{\circ }.$
(1)求$∠C$的度数;
解：$\because \angle BAC = 44^{\circ}$，
　　$\therefore \angle B + \angle C = 180^{\circ} - \angle BAC = 180^{\circ} - 44^{\circ} = 136^{\circ}$。
　　$\because \angle B = \angle C$，
　　$\therefore 2\angle C = 136^{\circ}$。
　　$\therefore \angle C =$$68^{\circ}$。

(2)若$∠ADE= 75^{\circ }$,求$∠CDE$的度数.
解：$\because \angle ADE = \angle AED$，$\angle ADE = 75^{\circ}$，
　　$\therefore \angle AED = 75^{\circ}$。
　$\because \angle AED + \angle CED = 180^{\circ}$，
　$\therefore \angle CED = 180^{\circ} - 75^{\circ} = 105^{\circ}$。
　　$\because \angle CDE + \angle CED + \angle C = 180^{\circ}$，
　$\therefore \angle CDE = 180^{\circ} - 105^{\circ} - 68^{\circ} =$$7^{\circ}$。

答案： 解:
(1)$\because \angle BAC = 44^{\circ}$，
　　$\therefore \angle B + \angle C = 180^{\circ} - \angle BAC = 180^{\circ} - 44^{\circ} = 136^{\circ}$。
　　$\because \angle B = \angle C$，
　　$\therefore 2\angle C = 136^{\circ}$。
　　$\therefore \angle C = 68^{\circ}$。
(2)$\because \angle ADE = \angle AED$，$\angle ADE = 75^{\circ}$，
　　$\therefore \angle AED = 75^{\circ}$。
　$\because \angle AED + \angle CED = 180^{\circ}$，
　$\therefore \angle CED = 180^{\circ} - 75^{\circ} = 105^{\circ}$。
　　$\because \angle CDE + \angle CED + \angle C = 180^{\circ}$，
　$\therefore \angle CDE = 180^{\circ} - 105^{\circ} - 68^{\circ} = 7^{\circ}$。

12.如图所示,在$△ABC$中,BE平分$∠ABD$,CE平分$∠ACD$,且$∠BEC= $$27^{\circ }$,求$∠BAC$的度数.

解:$\because \angle ABC$与$\angle ACD$的角平分线相交于点$E$，$\therefore \angle CBE =$$\frac{1}{2}\angle ABC$，$\angle ECD =$$\frac{1}{2}\angle ACD$。
　由三角形的外角性质得，$\angle ACD = \angle ABC + \angle BAC$，$\angle ECD = \angle BEC + \angle CBE$，
　$\therefore \frac{1}{2}\angle ACD = \angle BEC + \frac{1}{2}\angle ABC$。
　$\therefore \frac{1}{2}(\angle ABC + \angle BAC) = \angle BEC + \frac{1}{2}\angle ABC$，整理得，$\angle BAC =$$2\angle BEC$；
　　$\because \angle BEC = 27^{\circ}$，
　$\therefore \angle BAC =$$2× 27^{\circ} = 54^{\circ}$。

答案： 解:$\because \angle ABC$与$\angle ACD$的角平分线相交于点$E$，$\therefore \angle CBE = \frac{1}{2}\angle ABC$，$\angle ECD = \frac{1}{2}\angle ACD$。
　由三角形的外角性质得，$\angle ACD = \angle ABC + \angle BAC$，$\angle ECD = \angle BEC + \angle CBE$，
　$\therefore \frac{1}{2}\angle ACD = \angle BEC + \frac{1}{2}\angle ABC$。
　$\therefore \frac{1}{2}(\angle ABC + \angle BAC) = \angle BEC + \frac{1}{2}\angle ABC$，整理得，$\angle BAC = 2\angle BEC$；
　　$\because \angle BEC = 27^{\circ}$，
　$\therefore \angle BAC = 2× 27^{\circ} = 54^{\circ}$。

13.(1)如图1所示,AF与EB交于点O,我们将$△EOF与△AOB$称为对顶三角形.求证:$∠A+∠B= ∠E+∠F.$
(2)如图2所示,BE,CD相交于点A,$∠DEA,∠BCA$的平分线交于点F.求证:$∠F= \frac {1}{2}(∠B+∠D).$
(3)在(2)的条件下,$∠B:∠D:$
$∠F= 2:4:x$.求x.

答案：
解:
(1)由题意，知$\angle A + \angle B + \angle AOB = 180^{\circ}$，$\angle E + \angle F + \angle EOF = 180^{\circ}$，
　$\because \angle AOB = \angle EOF$，
　$\therefore \angle A + \angle B = \angle E + \angle F$。
(2)如图所示，由
(1)的结论，得$\angle F + \angle 3 = \angle D + \angle 1$，$\angle F + \angle 2 = \angle B + \angle 4$，
　$\therefore 2\angle F + \angle 3 + \angle 2 = \angle D + \angle B + \angle 1 + \angle 4$。
　$\because EF$，$CF$平分$\angle DEA$与$\angle BCA$，
　$\therefore \angle 1 = \angle 2$，$\angle 3 = \angle 4$。
　$\therefore 2\angle F = \angle D + \angle B$。
　$\therefore \angle F = \frac{1}{2}(\angle B + \angle D)$。
(3)设$\angle B = 2k(k \neq 0)$，则$\angle D = 4k$，$\angle F = xk$。
　由
(2)知$\angle F = \frac{1}{2}(\angle B + \angle D)$，即$xk = \frac{1}{2}(2k + 4k)$，解得$x = 3$。