《2025年1加1轻巧夺冠完美期末八年级数学上册人教版辽宁专版》

23. (13分)
如图,在平面直角坐标系$x O y$中,点$A$在$y$轴正半轴上,点$B$在第二象限,$\triangle A O B$为等边三角形,点$B$的横坐标为$-4$,过点$B$作$B C \perp A B$,交$y$轴于点$C$.
(1)如图1,求$B C$的长.
(2)如图2,$P$为$x$轴正半轴上一点,点$Q$在$B C$上,连接$A P,A Q$,使$\angle P A Q=60^{\circ}$,设点$P$的横坐标为$t(0＜t＜8)$,线段$C Q$的长为$n$,用含$t$的代数式表示$n$.
(3)如图3,在(2)的条件下,连接$O Q$,当$A Q+O Q$取最小值时,在平面直角坐标系$x O y$中取一点$D$,连接$D P,D Q$,使$D P=D Q$,且$\angle P D Q=3 \angle B A Q$,请直接

写

出

此时$t$的值及点$D$的坐标.

答案：
23.解：
(1)如答图1，过点$B$作$BE \perp y$轴于点$E$，则$\angle CEB = 90^{\circ}$，$\because$点$B$的横坐标为$-4$，$\therefore BE = 4$，$\because \triangle AOB$为等边三角形，$\therefore \angle BAO = 60^{\circ}$，$\because AB \perp BC$，$\therefore \angle ABC = 90^{\circ}$，$\therefore \angle BCE = 30^{\circ}$，$\therefore BC = 2BE = 8$.
　　　　　　　　
(2)$\because$点$P$的横坐标为$t(0 ＜ t ＜ 8)$，$\therefore OP = t$，$\because \triangle AOB$为等边三角形，$\therefore AB = AO$，$\angle BAO = 60^{\circ}$，$\because \angle BAO = \angle PAQ = 60^{\circ}$，$\therefore \angle BAQ = \angle OAP$，在$\triangle ABQ$和$\triangle AOP$中，$\begin{cases} \angle ABQ = \angle AOP, \\ AB = AO, \\ \angle BAQ = \angle OAP, \end{cases}$ $\therefore \triangle ABQ \cong \triangle AOP(ASA)$，$\therefore BQ = OP = t$，由
(1)知$BC = 8$，$\therefore CQ = 8 - t$，即$n = 8 - t$.
(3)如答图2，延长$AB$交$x$轴于点$A'$，$\because \triangle AOB$为等边三角形，$\therefore AB = BO = AO$，$\angle ABO = \angle AOB = 60^{\circ}$，$\because \angle AOA' = 90^{\circ}$，$\therefore \angle BOA' = \angle AA'O = 30^{\circ}$，$\therefore OB = A'B = AB$，$\because AB \perp BC$，$\therefore$点$A$与点$A'$关于$BC$对称，$BC$与$x$轴交于点$Q$，$\therefore$此时$AQ + OQ$的值最小，由
(2)知$BQ = OP = t$，$CQ = 8 - t$，$\because AB = AO$，$AQ = AQ$，$\angle ABQ = \angle AOQ = 90^{\circ}$，$\therefore Rt \triangle ABQ \cong Rt \triangle AOQ(HL)$，$\therefore OQ = BQ = t$，$\angle BAQ = \angle OAQ = 30^{\circ}$，$\angle OCQ = 30^{\circ}$，$\therefore OQ = \frac{1}{2}CQ$，$\therefore t = \frac{1}{2}(8 - t)$，解得$t = \frac{8}{3}$，$\because OQ = OP$，$PQ \perp y$轴，$\therefore y$轴是$PQ$的垂直平分线，$\because DP = DQ$，$\therefore$点$D$在$y$轴上，$\because \angle BAQ = 30^{\circ}$，$\angle PDQ = 3\angle BAQ$，$\therefore \angle PDQ = 90^{\circ}$，$\therefore \triangle PDQ$是等腰直角三角形，$\therefore OD = OQ = OP = \frac{8}{3}$，$\therefore$点$D$的坐标为$(0,\frac{8}{3})$或$(0,-\frac{8}{3})$.