在Rt?CHQ中，由勾股定理得：HQ?CQ2?CH2?52?42?3，则APQ?2HQ?6.………………………（9分） ②当点D在线段AM的延长线上时，∵?ABC与?DEC都是等边三角形 ∴AC?BC，CD?CE，?ACB??DCE?60? ∴?ACB??DCB??DCB??DCE ∴?ACD??BCE ∴?ACD≌?BCE?SAS? PBMC∴
?CBE??CAD?30?，同理可得：DQEPQ?6.…………………………（11分） ③当点D在线段MA的延长线上时， ∵?ABC与?DEC都是等边三角形 ∴AC?BC，CD?CE，?ACB??DCE?60? ∴?ACD??ACE??BCE??ACE?60? ∴?ACD??BCE ∴?ACD≌?BCE?SAS? ∴?CBE??CAD ∵?CAM?30?
∴?CBE??CAD?150? ∴?CBQ?30?. 同理可得：PQ?6.
综上，PQ的长是6. ………………………（13分）
DAEBMPCQ3．（2010 山东济南）（1）如图，已知AB?AC，AD?AE．求证BD?CE．
A B
D E
C
【答案】证明：∵AB=AC
∴∠B=∠C ∵AD=AE ∴∠ADE=∠AEC
∴180O -∠ADE=180O -∠AEC 即∠ADB=∠AEC 在△ABD和△ACE中
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∵AB=AC ∠B=∠C
∠ADB=∠AEC
∴△ABD≌△ACE
∴BD=CE
4．（2010湖南衡阳）已知：如图，在等边三角形ABC的AC边上取中点D，BC的延长线上取一点E，使 CE ＝ CD．求证：BD ＝ DE．
、【答案】∵△ABC是等边三角形，∴∠ABC=∠ACB=60°，∵D为AC中点，∴∠DBC=30°，∵CE ＝ CD，∴∠E=30°，∴∠DBC=∠E，∴BD ＝ DE.
5．（2010 山东省德州）如图，点E，F在BC上，BE＝CF，∠A＝∠D，∠B＝∠C，AF与DE交于点O． (1)求证：AB＝DC；
(2)试判断△OEF的形状，并说明理由．
A
O D
B E
第18题图
F C ＝CF，
A
O D
【答案】证明：（１）∵BE∴BE＋EF＝CF＋EF， 即BF＝CE．
又∵∠A＝∠D，∠B＝∠C，
∴△ABF≌△DCE（AAS）， ∴AB＝DC． （２）△OEF为等腰三角形 理由如下：∵△ABF≌△DCE， ∴∠AFB=∠DEC．
B E F C 第 14 页 共 19 页
∴OE=OF．
∴△OEF为等腰三角形．
6．（2010江苏常州）如图，在△ABC中，点D、E分别在边AC、AB上，BD=CE， ∠DBC=∠ECB。 求证：AB=AC。
【答案】
7．（2010四川内江）如图，△ACD和△BCE都是等腰直角三角形，∠ACD＝∠BCE＝90°，
AE交DC于F，BD分别交CE，AE于点G、H. 试猜测线段AE和BD的位置和数量关系，并说明理由.
DF H G ECA
【答案】解：猜测 AE＝BD，AE⊥BD. ··················································································· 2分 理由如下：
∵∠ACD＝∠BCE＝90°，
∴∠ACD＋∠DCE＝∠BCE＋∠DCE，即∠ACE＝∠DCB. ·················································· 3分 ∵△ACD和△BCE都是等腰直角三角形，
∴AC＝CD，CE＝CB. ············································································································ 4分 ∴△ACE≌△DCB（S.A.S.） ··································································································· 5分 ∴AE＝BD， ···························································································································· 6分 ∠CAE＝∠CDB,. ····················································································································· 7分 ∵∠AFC＝∠DFH，
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B
∴∠DHF＝∠ACD＝90°， ··································································································· 8分 ∴AE⊥BD. ··························································································································· 9分 8．（2010 福建三明）如图，?ACB和?BCD都是等腰直角三角形，∠ACB=∠ECD=90°，D为AB边上一点。 全品中考网
（1）求证：△ACE≌△BCD；（5分） （2）若AD=5，BD=12，求DE的长。（5分）
【答案】（1）证明：??ACB和?ECD都是等腰直角三角形
∴AC=BC，EC=DC
…………2分
??ACE??DCE??DCA,?BCD??ACB??DCA?ACB??ECD?90? …………3分 ??ACE??BCD
在?ACE和?BCD中，AC=BC EC=DC ?ACE??BCD
…………3分 ??ACE≌?BCD
（2）解：由（1）可得AE=BD，?EAC??DBC?45?
又?BAC?45?
??EAD??EAC??BAC?90?，即?EAD是直角三角形…………8分
?DE?AE2?AD2?13
…………10分
9．（2010湖北襄樊） 如图5，点E、C在BF上，BF=FC，∠ABC=∠DEF=45°，∠A=∠
D=90°． （1）求证：AB=DE；
（2）若AC交DE于M，且AB=3，ME=2，将线段CE绕点C顺时针旋转，使点E旋转到AB上的G处，求旋转角∠ECG的度数．
AGBEMCFD
图5
【答案】（1）∵BE=FC，∴BC=EF．
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