亚洲激情+欧美激情,无码任你躁久久久久久,我的极品美女老婆,性欧美牲交在线视频,亚洲av高清在线一区二区三区

答案：3

答案：
(1) 解：$BM + CN = MN$。
證明：如答圖①，延長$MB$至點$P$，使$BP = CN$，連接$DP$。
$\because \triangle ABC$為等邊三角形，
$\therefore \angle ACB = \angle ABC = 60^{\circ}$。
$\because BD = CD$，$\angle BDC = 120^{\circ}$，
$\therefore \angle DBC = \angle DCB = 30^{\circ}$，
$\therefore \angle ACD = \angle ABD = 90^{\circ} = \angle DBP$。
在$\triangle BDP$和$\triangle CDN$中，
$\left\{\begin{array}{l} BD = CD, \\ \angle DBP = \angle DCN, \\ BP = CN, \end{array}\right.$
$\therefore \triangle BDP \cong \triangle CDN(SAS)$。
$\therefore DP = DN$，$\angle BDP = \angle CDN$。
$\because \angle BDM + \angle CDN = 60^{\circ}$，
$\therefore \angle BDM + \angle BDP = 60^{\circ}$，即$\angle PDM = 60^{\circ}$，
$\therefore \angle PDM = \angle NDM$。
在$\triangle PDM$和$\triangle NDM$中，
$\left\{\begin{array}{l} DP = DN, \\ \angle PDM = \angle NDM, \\ DM = DM, \end{array}\right.$
$\therefore \triangle PDM \cong \triangle NDM(SAS)$。
$\therefore MP = MN$。$\therefore BM + CN = MN$。

(2) $2n + \frac{2}{3}l$ 點撥：過點$D$作$\angle CDH = \angle MDB$，邊$DH$交線段$AC$于點$H$，如答圖②。
由(1)知$\angle DCH = \angle MBD = 90^{\circ}$。
在$\triangle BMD$和$\triangle CHD$中，
$\left\{\begin{array}{l} \angle MBD = \angle HCD, \\ BD = CD, \\ \angle BDM = \angle CDH, \end{array}\right.$
$\therefore \triangle BMD \cong \triangle CHD(ASA)$，
$\therefore BM = CH$，$DM = DH$。
$\because \angle CDH = \angle MDB$，$\therefore \angle MDH = \angle BDC = 120^{\circ}$。
$\because \angle MDN = 60^{\circ}$，$\therefore \angle NDH = 120^{\circ} - 60^{\circ} = 60^{\circ}$，
$\therefore \angle MDN = \angle NDH$。
在$\triangle MDN$和$\triangle HDN$中，
$\left\{\begin{array}{l} DM = DH, \\ \angle MDN = \angle HDN, \\ ND = ND, \end{array}\right.$
$\therefore \triangle MDN \cong \triangle HDN(SAS)$，
$\therefore NM = NH = AN + AC - CH = AN + AC - BM$，
$\therefore \triangle AMN$的周長$= AN + AM + MN$
$= AN + AB + BM + AN + AC - BM$
$= 2AN + 2AB$。
$\because AN = n$，$AB = \frac{1}{3}l$，$\therefore \triangle AMN$的周長$= 2n + \frac{2}{3}l$。

答案：
(1) 證明：如答圖①，延長$EB$到點$G$，使$BG = DF$，連接$AG$。

$\because \angle ABG = \angle ABC = \angle D = 90^{\circ}$，$AB = AD$，
$\therefore \triangle ABG \cong \triangle ADF(SAS)$。
$\therefore AG = AF$，$\angle 1 = \angle 2$。
$\therefore \angle 1 + \angle 3 = \angle 2 + \angle 3 = \angle EAF = \frac{1}{2}\angle BAD$。
$\therefore \angle GAE = \angle EAF$。
又$\because AE = AE$，
$\therefore \triangle AEG \cong \triangle AEF$。
$\therefore EG = EF$。
$\because EG = BE + BG$，
$\therefore EF = BE + FD$。
(2) 解：(1)中的結(jié)論$EF = BE + FD$仍然成立。
(3) 解：結(jié)論$EF = BE + FD$不成立，應(yīng)當是$EF = BE - FD$。
證明：如答圖②，在$BE$上截取$BG$，使$BG = DF$，連接$AG$。

$\because \angle B + \angle ADC = 180^{\circ}$，$\angle ADF + \angle ADC = 180^{\circ}$，
$\therefore \angle B = \angle ADF$。
$\because AB = AD$，
$\therefore \triangle ABG \cong \triangle ADF$。
$\therefore \angle BAG = \angle DAF$，$AG = AF$。
$\therefore \angle BAG + \angle EAD = \angle DAF + \angle EAD$
$= \angle EAF = \frac{1}{2}\angle BAD$。
$\therefore \angle GAE = \angle EAF$。
$\because AE = AE$，
$\therefore \triangle AEG \cong \triangle AEF(SAS)$，
$\therefore EG = EF$。
$\because EG = BE - BG$，
$\therefore EF = BE - FD$。