《2025年暑假大串联七年级数学人教版》

21. 如图，直线$AB$，$CD$被$EF$所截，$\angle 1+\angle 2 = 180^{\circ}$，$EM$，$FN$分别平分$\angle BEF$和$\angle CFE$。
（1）试判定$EM$与$FN$之间的关系，并说明理由；
（2）由（1）的结论我们可以得到一个命题：如果两条直线______，那么内错角的平分线互相______；
（3）由此可以探究并得到：如果两条直线______，那么同旁内角的平分线互相______。

答案：
（1）$EM // FN$. 理由如下：$\because \angle 1 + \angle 2 = 180^{\circ}$，$\angle EFD + \angle 2 = 180^{\circ}$，$\therefore \angle 1 = \angle EFD$，$\therefore AB // CD$，$\therefore \angle BEF = \angle CFE$，$\because EM$，$FN$ 分别平分 $\angle BEF$ 和 $\angle CFE$，$\therefore \angle 3 = \angle 4$，$\therefore EM // FN$；
（2）由（1）的结论我们可以得到一个命题：如果两条直线平行，那么内错角的平分线互相平行，故答案为平行，平行；
（3）如图所示，假设 $EM$，$FM$ 分别平分 $\angle BEF$ 和 $\angle DFE$.

由（1）可得 $AB // CD$，$\therefore \angle BEF + \angle DFE = 180^{\circ}$，$\because EM$，$FM$ 分别平分 $\angle BEF$ 和 $\angle DFE$，$\therefore \angle MEF = \frac{1}{2}\angle BEF$，$\angle EFM = \frac{1}{2}\angle DFE$，$\therefore \angle MEF + \angle EFM = \frac{1}{2}(\angle BEF + \angle DFE) = 90^{\circ}$，$\therefore \angle EMF = 90^{\circ}$，$\therefore$ 由此可以探究并得到：如果两条直线平行，那么同旁内角的平分线互相垂直，故答案为平行，垂直.

22. 如图，$MN// BC$，$BD\perp DC$，$\angle 1=\angle 2 = 60^{\circ}$，$DC$是$\angle NDE$的平分线。
（1）$AB$与$DE$平行吗？请说明理由；
$AB // DE$. 理由如下：$\because MN // BC$，$\angle 1 = 60^{\circ}$，$\therefore \angle ABC = \angle 1 = 60^{\circ}$，又 $\because \angle 1 = \angle 2$，$\therefore \angle ABC = \angle 2 = 60^{\circ}$，$\therefore AB // DE$
（2）试说明$\angle ABC=\angle C$；
$\because MN // BC$，$\therefore \angle NDE + \angle 2 = 180^{\circ}$，而 $\angle 2 = 60^{\circ}$，$\therefore \angle NDE = 180^{\circ} - \angle 2 = 180^{\circ} - 60^{\circ} = 120^{\circ}$，$\because DC$ 是 $\angle NDE$ 的角平分线，$\therefore \angle EDC = \angle NDC = \frac{1}{2}\angle NDE = 60^{\circ}$，$\because MN // BC$，$\therefore \angle C = \angle NDC = 60^{\circ}$，$\therefore \angle ABC = \angle C = 60^{\circ}$
（3）求$\angle ABD$的度数。
$\because \angle ADC + \angle NDC = 180^{\circ}$，$\angle NDC = 60^{\circ}$，$\therefore \angle ADC = 180^{\circ} - \angle NDC = 180^{\circ} - 60^{\circ} = 120^{\circ}$，$\because BD \perp DC$，$\therefore \angle BDC = 90^{\circ}$，$\therefore \angle ADB = \angle ADC - \angle BDC = 120^{\circ} - 90^{\circ} = 30^{\circ}$，$\because MN // BC$，$\therefore \angle DBC = \angle ADB = 30^{\circ}$，$\because \angle ABC = \angle C = 60^{\circ}$，$\therefore \angle ABD = \angle ABC - \angle DBC = 30^{\circ}$

答案： （1）$AB // DE$. 理由如下：$\because MN // BC$，$\angle 1 = 60^{\circ}$，$\therefore \angle ABC = \angle 1 = 60^{\circ}$，又 $\because \angle 1 = \angle 2$，$\therefore \angle ABC = \angle 2 = 60^{\circ}$，$\therefore AB // DE$；
（2）$\because MN // BC$，$\therefore \angle NDE + \angle 2 = 180^{\circ}$，而 $\angle 2 = 60^{\circ}$，$\therefore \angle NDE = 180^{\circ} - \angle 2 = 180^{\circ} - 60^{\circ} = 120^{\circ}$，$\because DC$ 是 $\angle NDE$ 的角平分线，$\therefore \angle EDC = \angle NDC = \frac{1}{2}\angle NDE = 60^{\circ}$，$\because MN // BC$，$\therefore \angle C = \angle NDC = 60^{\circ}$，$\therefore \angle ABC = \angle C = 60^{\circ}$；
（3）$\because \angle ADC + \angle NDC = 180^{\circ}$，$\angle NDC = 60^{\circ}$，$\therefore \angle ADC = 180^{\circ} - \angle NDC = 180^{\circ} - 60^{\circ} = 120^{\circ}$，$\because BD \perp DC$，$\therefore \angle BDC = 90^{\circ}$，$\therefore \angle ADB = \angle ADC - \angle BDC = 120^{\circ} - 90^{\circ} = 30^{\circ}$，$\because MN // BC$，$\therefore \angle DBC = \angle ADB = 30^{\circ}$，$\because \angle ABC = \angle C = 60^{\circ}$，$\therefore \angle ABD = \angle ABC - \angle DBC = 30^{\circ}$.

23. 如图1，$\angle AOB = 40^{\circ}$，$OC$平分$\angle AOB$，点$E$在射线$OC$上，$ED\perp OA$，垂足为点$D$，$DF$平分$\angle ODE$，交射线$OC$于点$F$，点$P$是射线$OC$上一点，连接$DP$。
（1）如图1，若$DP$平分$\angle ODF$，则$\angle PDE=$______；
（2）如图2，若$DP// OB$，求$\angle PDE$的度数；
（3）如图3，若$DP\perp FD$，则$\angle ADP=$______；
（4）若$\angle PDF=\frac{2}{3}\angle EDF$，直接写出$\angle PDE$的度数。

答案：

（1）$\because ED \perp OA$，$\therefore \angle ODE = 90^{\circ}$，$\because DF$ 平分 $\angle ODE$，$\therefore \angle ODF = \angle EDF = \frac{1}{2}\angle ODE = 45^{\circ}$，$\because DP$ 平分 $\angle ODF$，$\therefore \angle PDF = \frac{1}{2}\angle ODF = 22.5^{\circ}$，$\therefore \angle PDE = \angle PDF + \angle EDF = 22.5^{\circ} + 45^{\circ} = 67.5^{\circ}$，故答案为 $67.5^{\circ}$；
（2）$\because DP // OB$，$\angle AOB = 40^{\circ}$，$\therefore \angle PDO + \angle AOB = 180^{\circ}$，$\therefore \angle PDO = 180^{\circ} - \angle AOB = 180^{\circ} - 40^{\circ} = 140^{\circ}$，$\because ED \perp OA$，$\therefore \angle ODE = 90^{\circ}$，$\therefore \angle PDE = \angle PDO - \angle ODE = 140^{\circ} - 90^{\circ} = 50^{\circ}$；
（3）$\because DP \perp FD$，$\therefore \angle FDP = 90^{\circ}$，$\because DF$ 平分 $\angle ODE$，$\angle ODE = 90^{\circ}$，$\therefore \angle ODF = \frac{1}{2}\angle ODE = 45^{\circ}$，$\therefore \angle ADP = 180^{\circ} - \angle ODF - \angle FDP = 180^{\circ} - 45^{\circ} - 90^{\circ} = 45^{\circ}$，故答案为 $45^{\circ}$；
（4）如图，当 $\angle PDF$ 在 $\angle EDF$ 的外部时，

$\because \angle EDF = 45^{\circ}$，$\angle PDF = \frac{2}{3}\angle EDF$，$\therefore \angle PDF = \frac{2}{3} \times 45^{\circ} = 30^{\circ}$，$\therefore \angle PDE = \angle PDF + \angle EDF = 30^{\circ} + 45^{\circ} = 75^{\circ}$；如图，当 $\angle PDF$ 在 $\angle EDF$ 的内部时，

同理，$\angle PDF = \frac{2}{3} \times 45^{\circ} = 30^{\circ}$，$\therefore \angle PDE = \angle EDF - \angle PDF = 45^{\circ} - 30^{\circ} = 15^{\circ}$；综上可知，$\angle PDE$ 的度数为 $75^{\circ}$ 或 $15^{\circ}$.