《2025年赢在假期期末加寒假八年级数学沪科版合肥工业大学出版社》

13. 如图所示，已知在平面直角坐标系中，点A的坐标为(0，2)，点B的坐标为(4，0)，AB的垂直平分线交x轴于点C，交AB于点D. 求△AOC的周长.

答案： 解：$\because AB$的垂直平分线交$x$轴于点$C$，$\therefore AC = BC$. $\because A(0,2)$，$B(4,0)$，$\therefore OA = 2$，$OB = 4$. $\therefore \triangle AOC$的周长$=OA + OC + AC = OA + OC + BC = OA + OB = 6$.

14. 如图所示，∠BAC = ∠ABD，AC = BD. 点O是AD，BC的交点，点E是AB的中点. 试判断OE和AB的位置关系，并给出证明.

答案： $OE\perp AB$. 证明：在$\triangle BAC$和$\triangle ABD$中，$\begin{cases}AC = BD,\\\angle BAC = \angle ABD,\\AB = BA,\end{cases}\therefore \triangle BAC\cong\triangle ABD$，$\therefore \angle OBA = \angle OAB$，$\therefore OA = OB$. 又$\because AE = BE$，$\therefore OE\perp AB$.

15. 如图所示，在△ABC中，AB = AC，∠BAC = 120°，D是BC的中点，DE⊥AB于点E. 求证：EB = 3EA.

答案： 证明：$\because AB = AC$，$\angle BAC = 120^{\circ}$，$\therefore \angle B = \angle C = 30^{\circ}$. 又$\because D$是$BC$的中点，$\therefore AD\perp BC$. 在$Rt\triangle ABD$中，$\angle B+\angle BAD = 90^{\circ}$，$\therefore \angle BAD = 60^{\circ}$. $\because DE\perp AB$，$\therefore \angle ADE+\angle DAE = 90^{\circ}$，$\therefore \angle ADE = 30^{\circ}$. 在$Rt\triangle ADE$中，$\angle ADE = 30^{\circ}$，在$Rt\triangle ABD$中，$\angle B = 30^{\circ}$. $\therefore EA=\frac{1}{2}AD$，$AD = \frac{1}{2}AB$，$\therefore EA=\frac{1}{4}AB$. 又$\because EA + EB = AB$，$\therefore EB=\frac{3}{4}AB$，$\therefore EB = 3EA$.