第33页 - 通城学典课时作业本八年级数学苏科版江苏专版

3

$30^{\circ}$

证明：∵$DE\perp AB，$$DF\perp BC，$∴$∠AED = ∠CF D = 90°$

∵$D$为$AC$的中点，∴$AD = CD$

$ $在$Rt\triangle AED$和$Rt\triangle CF D$中

$ \begin {cases}AD = CD\\DE = DF\end {cases}$

∴$Rt\triangle AED≌Rt\triangle CF D(\mathrm {HL})$

∴$∠A = ∠C，$∴$AB = BC$

∵$AB = AC，$∴$AB = BC = AC，$∴$\triangle ABC$是等边三角形

$ (1)$证明：∵$\triangle ABC，$$\triangle CDE$均为等边三角形

∴$AC = BC，$$CD = CE，$$∠ACB = ∠DCE = 60°$

∴$∠ACB + ∠BCE = ∠DCE + ∠BCE，$即$∠ACE = ∠BCD$

$ $在$\triangle ACE$和$\triangle BCD$中

$ \begin {cases}AC = BC\\∠ACE = ∠BCD\\CE = CD\end {cases}$

∴$\triangle ACE≌\triangle BCD(S AS)，$∴$AE = BD$

$ (2)$解：∵$\triangle ACE≌ \triangle BCD，$∴$∠CAE = ∠CBD$

又∵$\triangle AP C$与$\triangle BPO$的内角和均为$180°，$$∠AP C = ∠BPO$

∴$∠BOP = ∠ACP = 60°，$即$∠AOB = 60°$

证明：如图，过点$M$作$MQ// BC，$交$AC$于点$Q$

∵$\triangle ABC$是等边三角形

∴$∠A = ∠B = ∠ACB = 60°$

∵$MQ// BC$

∴$∠AMQ = ∠B = 60°，$$∠AQM = ∠ACB = 60°，$$∠QMP = ∠N$

∴$∠AMQ = ∠AQM = ∠A，$∴$\triangle AMQ $是等边三角形

∴$AM = QM$

∵$AM = CN，$∴$QM = CN$

$ $在$\triangle QMP $和$\triangle CNP {中}$

$ \begin {cases}∠QPM = ∠CPN \\∠QMP = ∠N\\QM = CN\end {cases}$

∴$\triangle QMP≌ \triangle CNP(\mathrm {AAS})$

∴$MP = NP$