1. (淄博中考)如图,在$\triangle ABC$中,$AB = AC$,$\angle A = 120^{\circ}$.分别以点$A和C$为圆心,以大于$\frac{1}{2}AC$的长度为半径作弧,两弧相交于点$P和点Q$,作直线$PQ分别交BC$,$AC于点D和点E$.若$CD = 3$,则$BD$的长为 ()

A. 4
B. 5
C. 6
D. 7

C 解析:连接$AD$，$\because AB = AC$，$\angle BAC = 120^{\circ}$，$\therefore \angle B = \angle C = 30^{\circ}$。由作法得$DE$垂直平分$AC$，$\therefore DA = DC = 3$，$\therefore \angle DAC = \angle C = 30^{\circ}$，$\therefore \angle BAD = 120^{\circ} - 30^{\circ} = 90^{\circ}$。在$Rt\triangle ABD$中，$\because \angle B = 30^{\circ}$，$\therefore BD = 2AD = 6$。故选 C。

2. (2024·扬州校级期中)如图,点$B为线段AQ$上的动点,$AQ = 4$,以$AB为边作等边\triangle ABC$,以$BC为底边作等腰三角形PCB$,则$PQ$的最小值为____.

2 解析:如图，连接$AP$，$\because \triangle ABC$是等边三角形，$\therefore AB = AC$，$\angle CAB = 60^{\circ}$。在$\triangle ABP$和$\triangle ACP$中，$\left\{\begin{array}{l} AB = AC,\\ BP = CP,\\ AP = AP,\end{array}\right.$ $\therefore \triangle ABP\cong \triangle ACP(SSS)$，$\therefore \angle CAP = \angle BAP$，$\therefore \angle PAQ = 30^{\circ}$，$\therefore$点$P$在射线$AP$上运动。当$QP\perp AP$时，$PQ$的值最小，$\therefore PQ = \frac{1}{2}AQ = \frac{1}{2}\times 4 = 2$。

3. 如图,在四边形$ABCD$中,$AD = 4$,$BC = 1$,$\angle A = 30^{\circ}$,$\angle B = 90^{\circ}$,$\angle ADC = 120^{\circ}$,则$CD$的长为____.

2 解析:如图，延长$AB$，$DC$交于点$E$，$\because \angle A = 30^{\circ}$，$\angle D = 120^{\circ}$，$\therefore \angle E = 30^{\circ}$，$\therefore \angle A = \angle E$，$\therefore AD = DE$。在$Rt\triangle BCE$中，$CE = 2BC = 2$，$\therefore CD = DE - CE = 4 - 2 = 2$。

4. 如图,$Rt\triangle ABC$中,$\angle ACB = 90^{\circ}$,$CA = CB$,$\angle BAD = \angle ADE = 60^{\circ}$,$DE = 3$,$AB = 10$,$CE平分\angle ACB$,$DE与CE相交于点E$,则$AD$的长为____.

7 解析:延长$DE$交$AB$于$F$，延长$CE$交$AB$于$G$，如图所示。$\because \angle BAD = \angle ADE = 60^{\circ}$，$\therefore AF = DF$，$\therefore \triangle ADF$是等边三角形，$\therefore AD = AF = DF$，$\angle AFD = 60^{\circ}$。$\because CA = CB$，$CE$平分$\angle ACB$，$\therefore CG\perp AB$，即$\angle CGB = 90^{\circ}$，$AG = \frac{1}{2}AB = 5$。设$AD = AF = DF = x$，在$Rt\triangle GEF$中，$\angle GFE = 60^{\circ}$，$\angle GEF = 30^{\circ}$，$EF = DF - DE = x - 3$，则$GF = \frac{1}{2}(x - 3)$。由$AF - GF = AG$得，$x - \frac{1}{2}(x - 3) = 5$，解得$x = 7$。

5. 如图,$\angle AOB = 60^{\circ}$,点$P在OA$上,$PC = PD$,若$OC = 5cm$,$OD = 8cm$,则$OP$的长是 ()

A. 13 cm
B. 12 cm
C. 8 cm
D. 5 cm

A 解析:过点$P$作$PE\perp OB$于点$E$，则$PE\perp CD$。$\because PC = PD$，$\therefore \triangle PCD$为等腰三角形，$\therefore$点$E$为$CD$的中点。$\because OC = 5cm$，$OD = 8cm$，$\therefore CD = 3cm$，$\therefore OE = 6.5cm$。$\because \angle AOB = 60^{\circ}$，$\therefore \angle OPE = 90^{\circ} - 60^{\circ} = 30^{\circ}$，$\therefore OP = 2OE = 13cm$，故选 A。

6. 如图,在$\triangle ABC$中,$AB = AC = 6$,$\angle B = \angle C = 15^{\circ}$,则$\triangle ABC$的面积为____.

9 解析:如图，延长$CA$，过点$B$作$BD\perp AC$于点$D$，$\because AC = AB = 6$，$\angle BAD = \angle ABC + \angle ACB = 30^{\circ}$，$\therefore BD = \frac{1}{2}AB = 3$，$\therefore \triangle ABC$的面积$ = \frac{1}{2}\times AC\cdot BD = \frac{1}{2}\times 6\times 3 = 9$。

7. 已知$CD是\triangle ABC$的高,$\angle BAC = 2\angle BCD$,$P是直线BC$上一点.
(1)当点$P在CB$的延长线上,且$\angle APC = 60^{\circ}$时,如图①,求证:$PB + PC = PA$.
(2)当点$P在边BC$上,且$\angle APC = 60^{\circ}$时,如图②;当点$P在边BC$上,且$\angle APC = 120^{\circ}$时,如图③,请直接写出线段$PB$,$PC$,$PA$之间的数量关系,不需要证明.

(1)如图①，过点$A$作$AH\perp BC$，垂足为$H$。$\because CD$是$\triangle ABC$的高，$\therefore \angle AHB = \angle AHC = \angle BDC = 90^{\circ}$，$\therefore \angle BAH + \angle ABC = 90^{\circ}$，$\angle BCD + \angle ABC = 90^{\circ}$，$\therefore \angle BAH = \angle BCD$。$\because \angle BAC = 2\angle BCD$，$\therefore \angle BAC = 2\angle BAH$，$\therefore \angle BAH = \angle CAH$。在$\triangle ABH$和$\triangle ACH$中，$\left\{\begin{array}{l} \angle AHB = \angle AHC,\\ AH = AH,\\ \angle BAH = \angle CAH,\end{array}\right.$ $\therefore \triangle ABH\cong \triangle ACH$，$\therefore BH = CH$。$\because \angle APC = 60^{\circ}$，$\therefore \angle PAH = 90^{\circ} - 60^{\circ} = 30^{\circ}$，$\therefore PA = 2PH$。$\because PB = PH - BH$，$PC = PH + HC$，$\therefore PB + PC = PH - BH + PH + CH = 2PH = PA$。
(2)题图②：$PC - PB = PA$。题图③：$PB - PC = PA$。解析:当点$P$在边$BC$上，且$\angle APC = 60^{\circ}$时，如图②，过点$A$作$AH\perp BC$，垂足为$H$。$\because CD$是$\triangle ABC$的高，$\therefore \angle AHB = \angle AHC = \angle BDC = 90^{\circ}$，$\therefore \angle BAH + \angle ABC = 90^{\circ}$，$\angle BCD + \angle ABC = 90^{\circ}$，$\therefore \angle BAH = \angle BCD$。$\because \angle BAC = 2\angle BCD$，$\therefore \angle BAC = 2\angle BAH$，$\therefore \angle BAH = \angle CAH$。在$\triangle ABH$和$\triangle ACH$中，$\left\{\begin{array}{l} \angle AHB = \angle AHC,\\ AH = AH,\\ \angle BAH = \angle CAH,\end{array}\right.$ $\therefore \triangle ABH\cong \triangle ACH$，$\therefore BH = CH$。$\because \angle APC = 60^{\circ}$，$\therefore \angle PAH = 90^{\circ} - 60^{\circ} = 30^{\circ}$，$\therefore PA = 2PH$。$\because PB = BH - PH$，$PC = PH + HC$，$\therefore PC - PB = PH + HC - BH + PH = 2PH = PA$，即$PC - PB = PA$。
当点$P$在边$BC$上，且$\angle APC = 120^{\circ}$时，如图③，过点$A$作$AH\perp BC$，垂足为$H$。$\because CD$是$\triangle ABC$的高，$\therefore \angle AHB = \angle AHC = \angle BDC = 90^{\circ}$，$\therefore \angle BAH + \angle ABC = 90^{\circ}$，$\angle BCD + \angle ABC = 90^{\circ}$，$\therefore \angle BAH = \angle BCD$。$\because \angle BAC = 2\angle BCD$，$\therefore \angle BAH = \angle CAH$。在$\triangle ABH$和$\triangle ACH$中，$\left\{\begin{array}{l} \angle AHB = \angle AHC,\\ AH = AH,\\ \angle BAH = \angle CAH,\end{array}\right.$ $\therefore \triangle ABH\cong \triangle ACH$，$\therefore BH = CH$。$\because \angle APC = 120^{\circ}$，$\therefore \angle APB = 60^{\circ}$，$\therefore \angle HAP = 30^{\circ}$，$\therefore PA = 2PH$。$\because PB = BH + PH$，$PC = HC - PH$，$\therefore PB - PC = BH + PH - HC + PH = 2PH = PA$，即$PB - PC = PA$。