三、解答题
16. 如图,点F,C在BD上,BF= CD,AB//DE,AC//EF.
(1)求证：△ABC≌△EDF；
(2)若AC= 6,∠ACB= 30°,求点E到BD的距离.
17. 如图,在△ABE中,∠ABE= 90°,∠E= 15°,AE的垂直平分线交BE于点C,连接AC,取AC的中点D.判断△ABD的形状,并说明理由.
18. 如图,D是等边三角形ABC的边AC上的一个定点,P,Q都是边AB上的动点(点P在点Q的上方),且PQ= AD.求证：CQ＞PD.
19. 如图,在△ABC中,∠C= 90°,AC= 10,BC= 5,AX⊥AC,垂足为A,点P和点Q分别在线段AC和射线AX上运动,且AB= PQ,当△ABC与△APQ全等时,请说明点P的位置.
20. 如图,在△ABC中,∠A= 40°,点D,E分别在边AB,AC上,BD= BC= CE,连接CD,BE.
(1)若∠ABC= 80°,求∠BDC,∠ABE的度数；
(2)写出∠BEC与∠BDC之间的数量关系,并说明理由.
21. 如图,在△ABC中,∠ABC,∠ACB的平分线交于点M,过点M作DE//BC,与边AB,AC分别交于点D,E.
(1)若∠A= 70°,求∠BMC的度数；
(2)若△ABC的周长为m,BC的长为a,求△ADE的周长(用含m,a的代数式表示).
22. 如图,已知Rt△ABC,∠A= 90°.
(1)用直尺和圆规完成下列问题：将△ABC绕点A按顺时针方向旋转90°得到△ADE(点B,C的对应点分别为D,E),BC的延长线交DE于点F,连接BD；(保留作图痕迹,不写作法)
(2)在(1)的条件下,若BC= 2EF,试确定△BDE的形状,并求∠ABC的大小.
答案：16.
(1) $\because AB// DE$，$\therefore \angle A = \angle D$．同理$\angle ACB = \angle DFE$．$\because BF = DC$，$\therefore BF + FC = DC + FC$，即$BC = DF$．$\therefore \triangle ABC\cong \triangle EDF$
(2) 作$EP\perp BD$，垂足为点$P$，$\because \triangle ABC\cong \triangle EDF$，$\therefore EF = AC = 6$．由
(1)知$\angle ACB = \angle DFE$，又$\angle ACB = 30^{\circ}$，$\therefore \angle DFE = 30^{\circ}$．$\therefore$ 在$Rt\triangle FEP$中，$EP = \frac{1}{2}EF = 3$，即点$E$到$BC$的距离为$3$（也可以用全等三角形对应边上的高相等）
17. $\triangle ABD$是等边三角形．证明：$\because AE$的垂直平分线交$BE$于点$C$，$\therefore CA = CE$．$\therefore \angle CAE = \angle E = 15^{\circ}$．$\therefore \angle ACB = \angle CAE + \angle E = 30^{\circ}$．$\therefore$ 在$Rt\triangle ABC$中，$\angle BAC = 60^{\circ}$．$\because$ 在$Rt\triangle ABC$中，$D$为$AC$的中点，$\therefore DA = DB$．$\therefore \triangle ABD$是等边三角形
18. 在$\triangle APD$中，$AD + AP ＞ PD$，$\because PQ = AD$，$\therefore PQ + AP ＞ PD$即$AQ ＞ PD$．$\because$ 点$Q$在$\triangle ABC$的边上，$\therefore \angle ACQ ＜ 60^{\circ}$．$\because \angle A = 60^{\circ}$，$\therefore \angle A ＞ \angle ACQ$．$\therefore CQ ＞ AQ$，$\therefore CQ ＞ PD$
19. $\because AX\perp AC$，$\therefore \angle C = \angle QAP = 90^{\circ}$．$\therefore$ 点$C$与点$A$是对应点．$\therefore$ 当$\triangle CAB\cong\triangle APQ$时，则$AP = CA = 10$，即$P$，$C$两点重合．$\therefore$ 当$\triangle CAB\cong\triangle AQP$时，则$AP = CB = 5$，即点$P$为$AC$中点．综上，当$\triangle ABC\cong\triangle APQ$时，点$P$与点$C$重合或为$AC$中点
20.
(1) 在$\triangle BDC$中，$BD = BC$，$\therefore \angle BDC = \angle BCD$．$\because \angle ABC = 80^{\circ}$，$\therefore \angle BDC = 50^{\circ}$．在$\triangle ABC$中，$\angle ABC = 80^{\circ}$，$\angle A = 40^{\circ}$，则$\angle ACB = 60^{\circ}$．$\because CE = CB$，$\therefore \triangle BCE$是等边三角形．$\therefore \angle BEC = 60^{\circ}$．$\because \angle BEC$是$\triangle ABE$的外角，$\therefore \angle ABE = \angle BEC - \angle A = 60^{\circ} - 40^{\circ} = 20^{\circ}$
(2) $\angle BDC + \angle BEC = 110^{\circ}$．理由：设$\angle ABC = \alpha^{\circ}$，在$\triangle BDC$中，$BD = BC$，则$\angle BDC = \frac{1}{2}(180 - \alpha)^{\circ}$．在$\triangle ABC$中，$\angle ACB = 180^{\circ} - 40^{\circ} - \alpha^{\circ} = (140 - \alpha)^{\circ}$．在$\triangle BEC$中，$CE = CB$，则$\angle BEC = \frac{1}{2}(180 - 140 + \alpha)^{\circ} = \frac{1}{2}(40 + \alpha)^{\circ}$．$\therefore \angle BDC + \angle BEC = \frac{1}{2}(180 - \alpha)^{\circ} + \frac{1}{2}(40 + \alpha)^{\circ} = 110^{\circ}$
21.
(1) $\because \angle ABC$，$\angle ACB$的平分线相交于点$M$，$\therefore \angle MBC = \frac{1}{2}\angle ABC$，$\angle MCB = \frac{1}{2}\angle ACB$．$\therefore \angle MBC + \angle MCB = \frac{1}{2}(\angle ABC + \angle ACB)$．在$\triangle ABC$中，$\angle A = 70^{\circ}$，则$\angle ABC + \angle ACB = 180^{\circ} - 70^{\circ} = 110^{\circ}$，$\therefore \angle MBC + \angle MCB = 55^{\circ}$．$\therefore \angle BMC = 125^{\circ}$
(2) $\because \angle ABC$，$\angle ACB$的平分线相交于点$M$，$\therefore \angle MBC = \angle DBM$，$\angle MCB = \angle ECM$．$\therefore DM = DB$，$EM = EC$．$\therefore \triangle ADE$的周长$ = AD + DE + AE = AB + AC = \triangle ABC$的周长$- BC = m - a$
22.
(1) 如图所示
(2) $\because \triangle ADE$是$\triangle ABC$绕顶点$A$按顺时针方向旋转$90^{\circ}$得到的，$\therefore \triangle ADE\cong\triangle ABC$．$\therefore \angle ABC = \angle ADE$，$BC = DE$，$AB = AD$．$\because \angle ADE + \angle DEA = 90^{\circ}$，$\therefore \angle ABC + \angle DEA = 90^{\circ}$．$\therefore BF\perp DE$．又$BC = 2EF$，$DE = 2EF$，即$BF$垂直平分$DE$，$\therefore BD = BE$，即$\triangle BDE$是等腰三角形．$\because AB = AD$，$\angle BAD = 90^{\circ}$，$\therefore$ 在等腰直角三角形$BDA$中，$\angle ABD = \angle BDA = 45^{\circ}$．$\because BD = BE$，$\therefore \angle BDE = \frac{1}{2}(180^{\circ} - 45^{\circ}) = 67.5^{\circ}$．$\therefore \angle ABC = 67.5^{\circ} - 45^{\circ} = 22.5^{\circ}$
