中考數(shù)學(xué)二輪培優(yōu)復(fù)習(xí)幾何專(zhuān)項(xiàng)練習(xí)：最值問(wèn)題之阿氏圓（含解析）

試卷第=page22頁(yè)，共=sectionpages22頁(yè)資料整理【淘寶店鋪：向陽(yáng)百分百】試卷第=page11頁(yè)，共=sectionpages11頁(yè)資料整理【淘寶店鋪：向陽(yáng)百分百】中考數(shù)學(xué)幾何專(zhuān)項(xiàng)練習(xí)：最值問(wèn)題之阿氏圓一、填空題1．如圖，正方形SKIPIF1<0的邊長(zhǎng)為4，SKIPIF1<0的半徑為2，SKIPIF1<0為SKIPIF1<0上的動(dòng)點(diǎn)，則SKIPIF1<0的最大值是．【答案】2【分析】解法1，如圖：以SKIPIF1<0為斜邊構(gòu)造等腰直角三角形SKIPIF1<0，連接SKIPIF1<0，SKIPIF1<0，連接SKIPIF1<0、SKIPIF1<0，推得SKIPIF1<0，因?yàn)镾KIPIF1<0，求出SKIPIF1<0即可求出答案．解法2：如圖：連接SKIPIF1<0、SKIPIF1<0、SKIPIF1<0，在SKIPIF1<0上做點(diǎn)SKIPIF1<0，使SKIPIF1<0，連接SKIPIF1<0，證明SKIPIF1<0SKIPIF1<0SKIPIF1<0，在SKIPIF1<0上做點(diǎn)SKIPIF1<0，使SKIPIF1<0，連接SKIPIF1<0，證明SKIPIF1<0SKIPIF1<0SKIPIF1<0，接著推導(dǎo)出SKIPIF1<0，最后證明SKIPIF1<0SKIPIF1<0SKIPIF1<0，即可求解．【詳解】解法1如圖：以SKIPIF1<0為斜邊構(gòu)造等腰直角三角形SKIPIF1<0，連接SKIPIF1<0，SKIPIF1<0，∴SKIPIF1<0，SKIPIF1<0，SKIPIF1<0四邊形SKIPIF1<0正方形SKIPIF1<0SKIPIF1<0，SKIPIF1<0又SKIPIF1<0SKIPIF1<0，SKIPIF1<0SKIPIF1<0SKIPIF1<0在SKIPIF1<0與SKIPIF1<0中SKIPIF1<0，SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0故答案為：2．解法2如圖：連接SKIPIF1<0、SKIPIF1<0、SKIPIF1<0根據(jù)題意正方形SKIPIF1<0的邊長(zhǎng)為4，SKIPIF1<0的半徑為2SKIPIF1<0SKIPIF1<0，SKIPIF1<0SKIPIF1<0SKIPIF1<0在SKIPIF1<0上做點(diǎn)SKIPIF1<0，使SKIPIF1<0，則SKIPIF1<0，連接SKIPIF1<0在SKIPIF1<0與SKIPIF1<0中SKIPIF1<0，SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0，則SKIPIF1<0SKIPIF1<0SKIPIF1<0在SKIPIF1<0上做點(diǎn)SKIPIF1<0，使SKIPIF1<0，則SKIPIF1<0，連接SKIPIF1<0在SKIPIF1<0與SKIPIF1<0中SKIPIF1<0，SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0，則SKIPIF1<0SKIPIF1<0如圖所示連接SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0在SKIPIF1<0與SKIPIF1<0中SKIPIF1<0，SKIPIF1<0，SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0故答案為：2．【點(diǎn)睛】本題考查正方形的性質(zhì)，相似三角形，勾股定理等知識(shí)，難度較大，熟悉以上知識(shí)點(diǎn)運(yùn)用是解題關(guān)鍵．2．如圖所示的平面直角坐標(biāo)系中，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0是第一象限內(nèi)一動(dòng)點(diǎn)，SKIPIF1<0，連接SKIPIF1<0、SKIPIF1<0，則SKIPIF1<0的最小值是．【答案】SKIPIF1<0【分析】取點(diǎn)SKIPIF1<0，連接SKIPIF1<0，SKIPIF1<0．根據(jù)SKIPIF1<0，有SKIPIF1<0，即可證明SKIPIF1<0，即有SKIPIF1<0，進(jìn)而可得SKIPIF1<0，則有SKIPIF1<0，利用勾股定理可得SKIPIF1<0，則有SKIPIF1<0，問(wèn)題得解．【詳解】解：如圖，取點(diǎn)SKIPIF1<0，連接SKIPIF1<0，SKIPIF1<0．SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，（當(dāng)B、P、T三點(diǎn)共線時(shí)取等號(hào)）SKIPIF1<0的最小值為SKIPIF1<0．故答案為：SKIPIF1<0．【點(diǎn)睛】本題考查阿氏圓問(wèn)題，相似三角形的判定和性質(zhì)，勾股定理等知識(shí)，解題的關(guān)鍵是學(xué)會(huì)添加常用輔助線，構(gòu)造相似三角形解決問(wèn)題．3．如圖所示，SKIPIF1<0，半徑為2的圓SKIPIF1<0內(nèi)切于SKIPIF1<0．SKIPIF1<0為圓SKIPIF1<0上一動(dòng)點(diǎn)，過(guò)點(diǎn)SKIPIF1<0作SKIPIF1<0、SKIPIF1<0分別垂直于SKIPIF1<0的兩邊，垂足為SKIPIF1<0、SKIPIF1<0，則SKIPIF1<0的取值范圍為．【答案】SKIPIF1<0【分析】根據(jù)題意，本題屬于動(dòng)點(diǎn)最值問(wèn)題-“阿氏圓”模型，首先作SKIPIF1<0于SKIPIF1<0，作SKIPIF1<0于SKIPIF1<0，如圖所示，通過(guò)代換，將SKIPIF1<0轉(zhuǎn)化為SKIPIF1<0，得到當(dāng)SKIPIF1<0與SKIPIF1<0相切時(shí)，SKIPIF1<0取得最大值和最小值，分兩種情況，作出圖形，數(shù)形結(jié)合解直角三角形即可得到相應(yīng)最值，進(jìn)而得到取值范圍．【詳解】解：作SKIPIF1<0于SKIPIF1<0，作SKIPIF1<0于SKIPIF1<0，如圖所示：SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0當(dāng)SKIPIF1<0與SKIPIF1<0相切時(shí)，SKIPIF1<0取得最大和最小，①連接SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，如圖1所示：可得：四邊形SKIPIF1<0是正方形，SKIPIF1<0，在SKIPIF1<0中，SKIPIF1<0，SKIPIF1<0，在SKIPIF1<0中，SKIPIF1<0，SKIPIF1<0，即SKIPIF1<0；②連接SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，如圖2所示：可得：四邊形SKIPIF1<0是正方形，SKIPIF1<0，由上同理可知：在SKIPIF1<0中，SKIPIF1<0，SKIPIF1<0，在SKIPIF1<0中，SKIPIF1<0，SKIPIF1<0，即SKIPIF1<0，SKIPIF1<0．故答案為：SKIPIF1<0．【點(diǎn)睛】本題考查動(dòng)點(diǎn)最值模型-“阿氏圓”，難度較大，掌握解決動(dòng)點(diǎn)最值問(wèn)題的方法，熟記相關(guān)幾何知識(shí)，尤其是圓的相關(guān)知識(shí)是解決問(wèn)題的關(guān)鍵．4．如圖，在SKIPIF1<0中，點(diǎn)A、點(diǎn)SKIPIF1<0在SKIPIF1<0上，SKIPIF1<0，SKIPIF1<0，點(diǎn)SKIPIF1<0在SKIPIF1<0上，且SKIPIF1<0，點(diǎn)SKIPIF1<0是SKIPIF1<0的中點(diǎn)，點(diǎn)SKIPIF1<0是劣弧SKIPIF1<0上的動(dòng)點(diǎn)，則SKIPIF1<0的最小值為．【答案】SKIPIF1<0【分析】延長(zhǎng)SKIPIF1<0到SKIPIF1<0，使得SKIPIF1<0，連接SKIPIF1<0，SKIPIF1<0，利用相似三角形的性質(zhì)證明SKIPIF1<0，求SKIPIF1<0的最小值問(wèn)題轉(zhuǎn)化為求SKIPIF1<0的最小值．求出SKIPIF1<0即可判斷．【詳解】解：延長(zhǎng)SKIPIF1<0到SKIPIF1<0，使得SKIPIF1<0，連接SKIPIF1<0，SKIPIF1<0．SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，又SKIPIF1<0在SKIPIF1<0中，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0的最小值為SKIPIF1<0，故答案為：SKIPIF1<0．【點(diǎn)睛】本題考查了相似三角形的判定和性質(zhì)，解直角三角形等知識(shí)，解題的關(guān)鍵是學(xué)會(huì)添加常用輔助線，構(gòu)造相似三角形解決問(wèn)題．5．如圖，邊長(zhǎng)為4的正方形，內(nèi)切圓記為⊙O，P是⊙O上一動(dòng)點(diǎn)，則SKIPIF1<0PA＋PB的最小值為．【答案】SKIPIF1<0【分析】SKIPIF1<0PA＋PB＝SKIPIF1<0（PA＋SKIPIF1<0PB），利用相似三角形構(gòu)造SKIPIF1<0PB即可解答．【詳解】解：設(shè)⊙O半徑為r，OP＝r＝SKIPIF1<0BC＝2，OB＝SKIPIF1<0r＝2SKIPIF1<0，取OB的中點(diǎn)I，連接PI，∴OI＝IB＝SKIPIF1<0，∵SKIPIF1<0，SKIPIF1<0，∴SKIPIF1<0，∠O是公共角，∴△BOP∽△POI，∴SKIPIF1<0，∴PI＝SKIPIF1<0PB，∴AP＋SKIPIF1<0PB＝AP＋PI，∴當(dāng)A、P、I在一條直線上時(shí)，AP＋SKIPIF1<0PB最小，作IE⊥AB于E，∵∠ABO＝45°，∴IE＝BE＝SKIPIF1<0BI＝1，∴AE＝AB?BE＝3，∴AI＝SKIPIF1<0，∴AP＋SKIPIF1<0PB最小值＝AI＝SKIPIF1<0，∵SKIPIF1<0PA＋PB＝SKIPIF1<0（PA＋SKIPIF1<0PB），∴SKIPIF1<0PA＋PB的最小值是SKIPIF1<0AI＝SKIPIF1<0．故答案是SKIPIF1<0．【點(diǎn)睛】本題是“阿氏圓”問(wèn)題，解決問(wèn)題的關(guān)鍵是構(gòu)造相似三角形．6．如圖，已知正方ABCD的邊長(zhǎng)為6，圓B的半徑為3，點(diǎn)P是圓B上的一個(gè)動(dòng)點(diǎn)，則SKIPIF1<0的最大值為．【答案】SKIPIF1<0【分析】如圖，連接SKIPIF1<0，在SKIPIF1<0上取一點(diǎn)SKIPIF1<0，使得SKIPIF1<0SKIPIF1<0SKIPIF1<0，進(jìn)而證明SKIPIF1<0,則在點(diǎn)P運(yùn)動(dòng)的任意時(shí)刻，均有PM=SKIPIF1<0，從而將問(wèn)題轉(zhuǎn)化為求PD-PM的最大值．連接PD，在△PDM中，PD-PM＜DM，故當(dāng)D、M、P共線時(shí)，PD-PM=DM為最大值，勾股定理即可求得SKIPIF1<0．【詳解】如圖，連接SKIPIF1<0，在SKIPIF1<0上取一點(diǎn)SKIPIF1<0，使得SKIPIF1<0SKIPIF1<0SKIPIF1<0，SKIPIF1<0，SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0在△PDM中，PD-PM＜DM，當(dāng)D、M、P共線時(shí)，PD-PM=DM為最大值，SKIPIF1<0四邊形SKIPIF1<0是正方形SKIPIF1<0在SKIPIF1<0中，SKIPIF1<0故答案為：SKIPIF1<0．【點(diǎn)睛】本題考查了圓的性質(zhì)，相似三角形的性質(zhì)與判定，勾股定理，構(gòu)造SKIPIF1<0是解題的關(guān)鍵．7．如圖，在邊長(zhǎng)為4的正方形ABCD內(nèi)有一動(dòng)點(diǎn)P，且BP＝SKIPIF1<0．連接CP，將線段PC繞點(diǎn)P逆時(shí)針旋轉(zhuǎn)90°得到線段PQ．連接CQ、DQ，則SKIPIF1<0DQ+CQ的最小值為．【答案】5【分析】連接AC、AQ，先證明△BCP∽△ACQ得SKIPIF1<0即AQ＝2，在AD上取AE＝1，證明△QAE∽△DAQ得EQ＝SKIPIF1<0QD，故SKIPIF1<0DQ+CQ＝EQ+CQ≥CE，求出CE即可．【詳解】解：如圖，連接AC、AQ，∵四邊形ABCD是正方形，PC繞點(diǎn)P逆時(shí)針旋轉(zhuǎn)90°得到線段PQ，∴∠ACB＝∠PCQ＝45°，∴∠BCP＝∠ACQ，cos∠ACB＝SKIPIF1<0，cos∠PCQ＝SKIPIF1<0，∴∠ACB=∠PCO，∴△BCP∽△ACQ，∴SKIPIF1<0∵BP＝SKIPIF1<0，∴AQ＝2，∴Q在以A為圓心，AQ為半徑的圓上，在AD上取AE＝1，∵SKIPIF1<0，SKIPIF1<0，∠QAE＝∠DAQ，∴△QAE∽△DAQ，∴SKIPIF1<0即EQ＝SKIPIF1<0QD，∴SKIPIF1<0DQ+CQ＝EQ+CQ≥CE，連接CE，∴SKIPIF1<0，∴SKIPIF1<0DQ+CQ的最小值為5．故答案為：5．【點(diǎn)睛】本題主要考查了正方形的性質(zhì)，旋轉(zhuǎn)的性質(zhì)，相似三角形的性質(zhì)與判定，三角函數(shù)，解題的關(guān)鍵在于能夠連接AC、AQ，證明兩對(duì)相似三角形求解.8．如圖，在SKIPIF1<0中，SKIPIF1<0，以點(diǎn)B為圓心作圓B與SKIPIF1<0相切，點(diǎn)P為圓B上任一動(dòng)點(diǎn)，則SKIPIF1<0的最小值是．【答案】SKIPIF1<0【分析】作BH⊥AC于H，取BC的中點(diǎn)D，連接PD，如圖，根據(jù)切線的性質(zhì)得BH為⊙B的半徑，再根據(jù)等腰直角三角形的性質(zhì)得到BHSKIPIF1<0ACSKIPIF1<0，接著證明△BPD∽△BCP得到PDSKIPIF1<0PC，所以PASKIPIF1<0PC＝PA+PD，而PA+PD≥AD（當(dāng)且僅當(dāng)A、P、D共線時(shí)取等號(hào)），從而計(jì)算出AD得到PASKIPIF1<0的最小值．【詳解】解：作BH⊥AC于H，取BC的中點(diǎn)D，連接PD，如圖，∵AC為切線，∴BH為⊙B的半徑，∵∠ABC＝90°，AB＝CB＝2，∴ACSKIPIF1<0BA＝2SKIPIF1<0，∴BHSKIPIF1<0ACSKIPIF1<0，∴BPSKIPIF1<0，∵SKIPIF1<0，SKIPIF1<0，而∠PBD＝∠CBP，∴△BPD∽△BCP，∴SKIPIF1<0，∴PDSKIPIF1<0PC，∴PASKIPIF1<0PC＝PA+PD，而PA+PD≥AD（當(dāng)且僅當(dāng)A、P、D共線時(shí)取等號(hào)），而ADSKIPIF1<0，∴PA+PD的最小值為SKIPIF1<0，即PASKIPIF1<0的最小值為SKIPIF1<0．故答案為：SKIPIF1<0．【點(diǎn)睛】本題考查了切線的性質(zhì)：圓的切線垂直于經(jīng)過(guò)切點(diǎn)的半徑．解決問(wèn)題的關(guān)鍵是利用相似比確定線段PDSKIPIF1<0PC．也考查了等腰直角三角形的性質(zhì)．9．如圖，在RtSKIPIF1<0中，AB＝AC＝4，點(diǎn)E，F(xiàn)分別是AB，AC的中點(diǎn)，點(diǎn)P是扇形AEF的SKIPIF1<0上任意一點(diǎn)，連接BP，CP，則SKIPIF1<0BP+CP的最小值是．【答案】SKIPIF1<0．【分析】在AB上取一點(diǎn)T，使得AT＝1，連接PT，PA，CT．證明SKIPIF1<0，推出SKIPIF1<0＝SKIPIF1<0＝SKIPIF1<0，推出PT＝SKIPIF1<0PB，推出SKIPIF1<0PB+CP＝CP+PT，根據(jù)PC+PT≥TC，求出CT即可解決問(wèn)題．【詳解】解：在AB上取一點(diǎn)T，使得AT＝1，連接PT，PA，CT．∵PA＝2．AT＝1，AB＝4，∴PA2＝SKIPIF1<0AT?AB，∴SKIPIF1<0＝SKIPIF1<0，∵∠PAT＝∠PAB，∴SKIPIF1<0，∴SKIPIF1<0＝SKIPIF1<0＝SKIPIF1<0，∴PT＝SKIPIF1<0PB，∴SKIPIF1<0PB+CP＝CP+PT，∵PC+PT≥TC，在RtSKIPIF1<0中，∵∠CAT＝90°，AT＝1，AC＝4，∴CT＝SKIPIF1<0＝SKIPIF1<0，∴SKIPIF1<0PB+PC≥SKIPIF1<0，∴SKIPIF1<0PB+PC的最小值為SKIPIF1<0．故答案為SKIPIF1<0．【點(diǎn)睛】本題考查等腰直角三角形的性質(zhì)，三角形相似的判定與性質(zhì)，勾股定理的應(yīng)用，三角形的三邊關(guān)系，圓的基本性質(zhì)，掌握以上知識(shí)是解題的關(guān)鍵．10．如圖，在△ABC中，∠ACB=90°，BC=12，AC=9，以點(diǎn)C為圓心，6為半徑的圓上有一個(gè)動(dòng)點(diǎn)D.連接AD、BD、CD，則2AD+3BD的最小值是.

【答案】SKIPIF1<0【分析】如下圖，在CA上取一點(diǎn)E，使得CE=4，先證△DCE∽△ACD，將SKIPIF1<0轉(zhuǎn)化為DE，從而求得SKIPIF1<0的最小距離，進(jìn)而得出2AD+3BD的最小值．【詳解】如下圖，在CA上取一點(diǎn)E，使得CE=4

∵AC=9，CD=6，CE=4∴SKIPIF1<0∵∠ECD=∠ACD∴△DCE∽△ACD∴SKIPIF1<0∴ED=SKIPIF1<0在△EDB中，ED+DB≥EB∴ED+DB最小為EB，即ED+DB=EB∴SKIPIF1<0在Rt△ECB中，EB=SKIPIF1<0∴SKIPIF1<0∴2AD+3DB=SKIPIF1<0故答案為：SKIPIF1<0．【點(diǎn)睛】本題考查求最值問(wèn)題，解題關(guān)鍵是構(gòu)造出△DCE∽△ACD．11．如圖，已知正方形ABCD的邊長(zhǎng)為4，⊙B的半徑為2，點(diǎn)P是⊙B上的一個(gè)動(dòng)點(diǎn)，則PD﹣SKIPIF1<0PC的最大值為．【答案】5【詳解】分析:由PD?SKIPIF1<0PC＝PD?PG≤DG，當(dāng)點(diǎn)P在DG的延長(zhǎng)線上時(shí)，PD?SKIPIF1<0PC的值最大，最大值為DG＝5．詳解:在BC上取一點(diǎn)G，使得BG＝1，如圖，∵SKIPIF1<0，SKIPIF1<0，∴SKIPIF1<0，∵∠PBG＝∠PBC，∴△PBG∽△CBP，∴SKIPIF1<0，∴PG＝SKIPIF1<0PC，當(dāng)點(diǎn)P在DG的延長(zhǎng)線上時(shí)，PD?SKIPIF1<0PC的值最大，最大值為DG＝SKIPIF1<0＝5．故答案為5點(diǎn)睛:本題考查圓綜合題、正方形的性質(zhì)、相似三角形的判定和性質(zhì)等知識(shí)，解題的關(guān)鍵是學(xué)會(huì)構(gòu)建相似三角形解決問(wèn)題，學(xué)會(huì)用轉(zhuǎn)化的思想思考問(wèn)題，把問(wèn)題轉(zhuǎn)化為兩點(diǎn)之間線段最短解決，題目比較難，屬于中考?jí)狠S題．二、解答題12．已知SKIPIF1<0與SKIPIF1<0有公共頂點(diǎn)C，SKIPIF1<0為等邊三角形，在SKIPIF1<0中，SKIPIF1<0．(1)如圖1，當(dāng)點(diǎn)E與點(diǎn)B重合時(shí)，連接AD，已知四邊形ABDC的面積為SKIPIF1<0，求SKIPIF1<0的值；(2)如圖2，SKIPIF1<0，A、E、D三點(diǎn)共線，連接SKIPIF1<0、SKIPIF1<0，取SKIPIF1<0中點(diǎn)M，連接SKIPIF1<0，求證：SKIPIF1<0；(3)如圖3，SKIPIF1<0，SKIPIF1<0，將SKIPIF1<0以C為旋轉(zhuǎn)中心旋轉(zhuǎn)，取SKIPIF1<0中點(diǎn)F，當(dāng)SKIPIF1<0的值最小時(shí)，求SKIPIF1<0的值．【答案】(1)SKIPIF1<0(2)見(jiàn)解析(3)SKIPIF1<0【分析】（1）延長(zhǎng)SKIPIF1<0到T，使得SKIPIF1<0連接SKIPIF1<0，過(guò)點(diǎn)D做SKIPIF1<0于N，證明SKIPIF1<0，得出SKIPIF1<0，SKIPIF1<0，證明SKIPIF1<0為等邊三角形，設(shè)SKIPIF1<0，得出SKIPIF1<0，求出x的值即可得出答案；（2）延長(zhǎng)SKIPIF1<0到SKIPIF1<0使得SKIPIF1<0，連接SKIPIF1<0、SKIPIF1<0，證明SKIPIF1<0，得出SKIPIF1<0，證明SKIPIF1<0為SKIPIF1<0的中位線，得出SKIPIF1<0，即可證明結(jié)論；（3）連接SKIPIF1<0，過(guò)點(diǎn)A作SKIPIF1<0于點(diǎn)G，以點(diǎn)C為圓心，SKIPIF1<0為半徑作圓，在SKIPIF1<0上截取SKIPIF1<0，連接SKIPIF1<0，證明SKIPIF1<0，得出SKIPIF1<0，即SKIPIF1<0，得出SKIPIF1<0，連接SKIPIF1<0與SKIPIF1<0交于一點(diǎn)，當(dāng)點(diǎn)F在此點(diǎn)時(shí)，SKIPIF1<0最小，即SKIPIF1<0最小，過(guò)點(diǎn)M作SKIPIF1<0于點(diǎn)N，過(guò)點(diǎn)A作SKIPIF1<0于點(diǎn)Q，求出SKIPIF1<0，SKIPIF1<0即可得出答案．【詳解】（1）解：延長(zhǎng)SKIPIF1<0到T，使得SKIPIF1<0連接SKIPIF1<0，過(guò)點(diǎn)D做SKIPIF1<0于N，如圖所示：∵SKIPIF1<0為等邊三角形，SKIPIF1<0，∴SKIPIF1<0，SKIPIF1<0，四邊形SKIPIF1<0中，SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0，在SKIPIF1<0和SKIPIF1<0中，SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0，SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0為等邊三角形，∵四邊形ABDC的面積為SKIPIF1<0，∴SKIPIF1<0，∵SKIPIF1<0，∴SKIPIF1<0，設(shè)SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0，（2）證明：延長(zhǎng)SKIPIF1<0到SKIPIF1<0使得SKIPIF1<0，連接SKIPIF1<0、SKIPIF1<0，如圖所示：∵SKIPIF1<0，SKIPIF1<0，∴SKIPIF1<0，∵SKIPIF1<0，∴SKIPIF1<0為等邊三角形，∴SKIPIF1<0，SKIPIF1<0，∵SKIPIF1<0為等邊三角形，∴SKIPIF1<0，SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0，在SKIPIF1<0和SKIPIF1<0中，SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0，∵A為SKIPIF1<0中點(diǎn)，M為SKIPIF1<0中點(diǎn)，∴SKIPIF1<0為SKIPIF1<0的中位線，∴SKIPIF1<0，∴SKIPIF1<0；（3）解：如圖，連接SKIPIF1<0，過(guò)點(diǎn)A作SKIPIF1<0于點(diǎn)G，以點(diǎn)C為圓心，SKIPIF1<0為半徑作圓，在SKIPIF1<0上截取SKIPIF1<0，連接SKIPIF1<0，∵SKIPIF1<0，SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0，SKIPIF1<0，∵SKIPIF1<0，SKIPIF1<0，∴SKIPIF1<0，∵點(diǎn)F為等邊三角形SKIPIF1<0的邊SKIPIF1<0中點(diǎn)，∴SKIPIF1<0，∴SKIPIF1<0，∵SKIPIF1<0，SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0，∵SKIPIF1<0的長(zhǎng)度為定值，∴在SKIPIF1<0旋轉(zhuǎn)時(shí)，點(diǎn)F在以C為圓心，SKIPIF1<0為半徑的圓上運(yùn)動(dòng)，∴如圖，連接SKIPIF1<0與SKIPIF1<0交于一點(diǎn)，當(dāng)點(diǎn)F在此點(diǎn)時(shí)，SKIPIF1<0最小，即SKIPIF1<0最小，過(guò)點(diǎn)M作SKIPIF1<0于點(diǎn)N，過(guò)點(diǎn)A作SKIPIF1<0于點(diǎn)Q，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，∴SKIPIF1<0，∵SKIPIF1<0，SKIPIF1<0，∴SKIPIF1<0，SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0，∵SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0．【點(diǎn)睛】本題主要考查了等邊三角形的判定和性質(zhì)，三角形全等的判定和性質(zhì)，三角形相似的判定和性質(zhì)，特殊角的三角函數(shù)，求正切值，勾股定理，直角三角形的性質(zhì)，解題的關(guān)鍵是作出輔助線，找出SKIPIF1<0取最小值時(shí)，點(diǎn)F的位置．13．如圖1，拋物線SKIPIF1<0與SKIPIF1<0軸交于SKIPIF1<0兩點(diǎn)，與SKIPIF1<0軸交于點(diǎn)SKIPIF1<0，其中點(diǎn)SKIPIF1<0的坐標(biāo)為SKIPIF1<0，拋物線的對(duì)稱(chēng)軸是直線SKIPIF1<0．(1)求拋物線的解析式；(2)若點(diǎn)SKIPIF1<0是直線SKIPIF1<0下方的拋物線上一個(gè)動(dòng)點(diǎn)，是否存在點(diǎn)SKIPIF1<0使四邊形SKIPIF1<0的面積為16，若存在，求出點(diǎn)SKIPIF1<0的坐標(biāo)若不存在，請(qǐng)說(shuō)明理由；(3)如圖2，過(guò)點(diǎn)SKIPIF1<0作SKIPIF1<0交拋物線的對(duì)稱(chēng)軸于點(diǎn)SKIPIF1<0，以點(diǎn)SKIPIF1<0為圓心，2為半徑作SKIPIF1<0，點(diǎn)SKIPIF1<0為SKIPIF1<0上的一個(gè)動(dòng)點(diǎn)，求SKIPIF1<0的最小值．【答案】(1)SKIPIF1<0(2)SKIPIF1<0或SKIPIF1<0(3)SKIPIF1<0【分析】（1）根據(jù)點(diǎn)SKIPIF1<0的坐標(biāo)為SKIPIF1<0，拋物線的對(duì)稱(chēng)軸是直線SKIPIF1<0．待定系數(shù)法求二次函數(shù)解析式即可，（2）先求得直線SKIPIF1<0解析式，設(shè)SKIPIF1<0，則SKIPIF1<0，過(guò)點(diǎn)SKIPIF1<0作SKIPIF1<0軸交直線SKIPIF1<0于點(diǎn)SKIPIF1<0，根據(jù)SKIPIF1<0等于16建立方程，解一元二次方程即可求得SKIPIF1<0的值，然后求得SKIPIF1<0的坐標(biāo)，（3）在SKIPIF1<0上取SKIPIF1<0，過(guò)點(diǎn)SKIPIF1<0作SKIPIF1<0，構(gòu)造SKIPIF1<0，則當(dāng)SKIPIF1<0三點(diǎn)共線時(shí)，取得最小值，最小值為SKIPIF1<0，勾股定理解直角三形即可．【詳解】（1）解：∵拋物線SKIPIF1<0與SKIPIF1<0軸交于SKIPIF1<0兩點(diǎn)，與SKIPIF1<0軸交于點(diǎn)SKIPIF1<0，點(diǎn)SKIPIF1<0的坐標(biāo)為SKIPIF1<0，拋物線的對(duì)稱(chēng)軸是直線SKIPIF1<0，∴SKIPIF1<0，SKIPIF1<0，解得SKIPIF1<0，SKIPIF1<0拋物線解析式為：SKIPIF1<0，（2）當(dāng)SKIPIF1<0，即SKIPIF1<0，解得SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，設(shè)直線SKIPIF1<0解析式為SKIPIF1<0，SKIPIF1<0，解得SKIPIF1<0，SKIPIF1<0直線SKIPIF1<0解析式為SKIPIF1<0，設(shè)SKIPIF1<0，過(guò)點(diǎn)SKIPIF1<0作SKIPIF1<0軸交直線SKIPIF1<0于點(diǎn)SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0SKIPIF1<0，SKIPIF1<0四邊形SKIPIF1<0的面積為16，SKIPIF1<0SKIPIF1<0SKIPIF1<0，解得SKIPIF1<0，SKIPIF1<0或SKIPIF1<0，（3）如圖，過(guò)點(diǎn)SKIPIF1<0作SKIPIF1<0交拋物線的對(duì)稱(chēng)軸于點(diǎn)SKIPIF1<0，以點(diǎn)SKIPIF1<0為圓心，2為半徑作SKIPIF1<0，SKIPIF1<0是拋物線的對(duì)稱(chēng)軸，SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，在SKIPIF1<0上取SKIPIF1<0，過(guò)點(diǎn)SKIPIF1<0作SKIPIF1<0，交SKIPIF1<0軸于點(diǎn)SKIPIF1<0，交拋物線對(duì)稱(chēng)軸于點(diǎn)SKIPIF1<0，則SKIPIF1<0SKIPIF1<0，SKIPIF1<0SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0SKIPIF1<0，當(dāng)SKIPIF1<0三點(diǎn)共線時(shí)，取得最小值，最小值為SKIPIF1<0，SKIPIF1<0SKIPIF1<0．則SKIPIF1<0的最小值為SKIPIF1<0．【點(diǎn)睛】本題考查了二次函數(shù)綜合，相似三角形的性質(zhì)與判定，掌握二次函數(shù)的性質(zhì)與相似三角形的性質(zhì)與判定是解題的關(guān)鍵．14．如圖1，拋物線SKIPIF1<0與x軸交于點(diǎn)SKIPIF1<0，與y軸交于點(diǎn)B，在x軸上有一動(dòng)點(diǎn)SKIPIF1<0（SKIPIF1<0），過(guò)點(diǎn)E作x軸的垂線交直線AB于點(diǎn)N，交拋物線于點(diǎn)P，過(guò)點(diǎn)P作PM⊥AB于點(diǎn)M．(1)求a的值和直線AB的函數(shù)表達(dá)式：(2)設(shè)△PMN的周長(zhǎng)為SKIPIF1<0，△AEN的周長(zhǎng)為SKIPIF1<0，若SKIPIF1<0求m的值．(3)如圖2，在（2）的條件下，將線段OE繞點(diǎn)O逆時(shí)針旋轉(zhuǎn)得到SKIPIF1<0，旋轉(zhuǎn)角為SKIPIF1<0（SKIPIF1<0），連接SKIPIF1<0、SKIPIF1<0，求SKIPIF1<0的最小值．【答案】(1)a=-SKIPIF1<0．直線AB解析式為y=-SKIPIF1<0x+3；(2)2(3)SKIPIF1<0【分析】（1）令y=0，求出拋物線與x軸交點(diǎn)，列出方程即可求出a，根據(jù)待定系數(shù)法可以確定直線AB解析式；（2）由△PNM∽△ANE，推出SKIPIF1<0，列出方程即可解決問(wèn)題；（3）在y軸上取一點(diǎn)M使得OM′=SKIPIF1<0，構(gòu)造相似三角形，可以證明AM′就是E′A+SKIPIF1<0E′B的最小值．【詳解】（1）令y=0，則ax2+（a+3）x+3=0，∴（x+1）（ax+3）=0，∴x=-1或-SKIPIF1<0，∵拋物線y=ax2+（a+3）x+3（a≠0）與x軸交于點(diǎn)A（4，0），∴-SKIPIF1<0=4，∴a=-SKIPIF1<0．∵A（4，0），B（0，3），設(shè)直線AB解析式為y=kx+b，則SKIPIF1<0，解得SKIPIF1<0，∴直線AB解析式為y=-SKIPIF1<0x+3；（2）如圖1，∵PM⊥AB，PE⊥OA，∴∠PMN=∠AEN，∵∠PNM=∠ANE，∴△PNM∽△ANE，∵SKIPIF1<0∴SKIPIF1<0，∵NE∥OB，∴SKIPIF1<0，∴SKIPIF1<0，∵拋物線解析式為SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0，解得m=2或4，經(jīng)檢驗(yàn)x=4是分式方程的增根，∴m=2；（3）如圖2，在y軸上取一點(diǎn)M′使得OM′=SKIPIF1<0，連接AM′，在AM′上取一點(diǎn)E′使得OE′=OE．∵OE′=2，OM′?OB=SKIPIF1<0，∴OE′2=OM′?OB，∴SKIPIF1<0，∵∠BOE′=∠M′OE′，∴△M′OE′∽△E′OB，∴SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0，此時(shí)SKIPIF1<0最?。▋牲c(diǎn)間線段最短，A、M′、E′共線時(shí)），最小值SKIPIF1<0．【點(diǎn)睛】本題為二次函數(shù)綜合題，主要考查相似三角形的判定和性質(zhì)、待定系數(shù)法、最小值問(wèn)題等知識(shí)，解題的關(guān)鍵是構(gòu)造相似三角形，找到線段AM′就是SKIPIF1<0的最小值．15．如圖，Rt△ABC，∠ACB＝90°，AC＝BC＝2，以C為頂點(diǎn)的正方形CDEF（C、D、E、F四個(gè)頂點(diǎn)按逆時(shí)針?lè)较蚺帕校┛梢岳@點(diǎn)C自由轉(zhuǎn)動(dòng)，且CD＝SKIPIF1<0，連接AF，BD（1）求證：△BDC≌△AFC（2）當(dāng)正方形CDEF有頂點(diǎn)在線段AB上時(shí)，直接寫(xiě)出BD＋SKIPIF1<0AD的值；（3）直接寫(xiě)出正方形CDEF旋轉(zhuǎn)過(guò)程中，BD＋SKIPIF1<0AD的最小值．【答案】（1）見(jiàn)解析；（2）SKIPIF1<0或SKIPIF1<0；（3）SKIPIF1<0【分析】（1）利用SAS，即可證明△FCA≌△DCB；（2）分兩種情況當(dāng)點(diǎn)D，E在AB邊上時(shí)和當(dāng)點(diǎn)E，F(xiàn)在邊AB上時(shí)，討論即可求解；（3）取AC的中點(diǎn)M．連接DM，BM．則CM＝1，可證得△DCM∽△ACD，可得DM＝SKIPIF1<0AD，從而得到當(dāng)B，D，M共線時(shí)，BD+SKIPIF1<0AD的值最小，即可求解．【詳解】（1）證明：∵四邊形CDEF是正方形，∴CF＝CD，∠DCF＝∠ACB＝90°，∴∠ACF＝∠DCB，∵AC＝CB，∴△FCA≌△DCB（SAS）；（2）解：①如圖2中，當(dāng)點(diǎn)D，E在AB邊上時(shí)，∵AC＝BC＝2，∠ACB＝90°，∴SKIPIF1<0，∵CD⊥AB，∴AD＝BD＝SKIPIF1<0，∴BD+SKIPIF1<0AD＝SKIPIF1<0；②如圖3中，當(dāng)點(diǎn)E，F(xiàn)在邊AB上時(shí)．BD＝CF＝SKIPIF1<0SKIPIF1<0，AD＝SKIPIF1<0＝SKIPIF1<0，∴BD+SKIPIF1<0AD＝SKIPIF1<0，綜上所述，BD＋SKIPIF1<0AD的值SKIPIF1<0或SKIPIF1<0；（3）如圖4中．取AC的中點(diǎn)M．連接DM，BM．則CM＝1，∵CD＝SKIPIF1<0，CM＝1，CA＝2，∴CD2＝CM?CA，∴SKIPIF1<0＝SKIPIF1<0，∵∠DCM＝∠ACD，∴△DCM∽△ACD，∴SKIPIF1<0＝SKIPIF1<0＝SKIPIF1<0，∴DM＝SKIPIF1<0AD，∴BD+SKIPIF1<0AD＝BD+DM，∴當(dāng)B，D，M共線時(shí)，BD+SKIPIF1<0AD的值最小，最小值SKIPIF1<0．【點(diǎn)睛】本題主要考查了相似三角形的判定和性質(zhì)，全等三角形的判定和性質(zhì)，正方形的性質(zhì)，銳角三角函數(shù)，熟練掌握相關(guān)知識(shí)點(diǎn)是解題的關(guān)鍵．16．如圖1，在平面直角坐標(biāo)系中，直線y＝﹣5x+5與x軸，y軸分別交于A，C兩點(diǎn)，拋物線y＝x2+bx+c經(jīng)過(guò)A，C兩點(diǎn)，與x軸的另一交點(diǎn)為B（1）求拋物線解析式及B點(diǎn)坐標(biāo)；（2）若點(diǎn)M為x軸下方拋物線上一動(dòng)點(diǎn)，連接MA、MB、BC，當(dāng)點(diǎn)M運(yùn)動(dòng)到某一位置時(shí)，四邊形AMBC面積最大，求此時(shí)點(diǎn)M的坐標(biāo)及四邊形AMBC的面積；（3）如圖2，若P點(diǎn)是半徑為2的⊙B上一動(dòng)點(diǎn)，連接PC、PA，當(dāng)點(diǎn)P運(yùn)動(dòng)到某一位置時(shí)，PC+SKIPIF1<0PA的值最小，請(qǐng)求出這個(gè)最小值，并說(shuō)明理由．【答案】（1）y＝x2﹣6x+5，B（5，0）；（2）當(dāng)M（3，﹣4）時(shí)，四邊形AMBC面積最大，最大面積等于18；（3）PC+SKIPIF1<0PA的最小值為SKIPIF1<0，理由詳見(jiàn)解析.【分析】（1）由直線y＝﹣5x+5求點(diǎn)A、C坐標(biāo)，用待定系數(shù)法求拋物線解析式，進(jìn)而求得點(diǎn)B坐標(biāo)．（2）從x軸把四邊形AMBC分成△ABC與△ABM；由點(diǎn)A、B、C坐標(biāo)求△ABC面積；設(shè)點(diǎn)M橫坐標(biāo)為m，過(guò)點(diǎn)M作x軸的垂線段MH，則能用m表示MH的長(zhǎng)，進(jìn)而求△ABM的面積，得到△ABM面積與m的二次函數(shù)關(guān)系式，且對(duì)應(yīng)的a值小于0，配方即求得m為何值時(shí)取得最大值，進(jìn)而求點(diǎn)M坐標(biāo)和四邊形AMBC的面積最大值．（3）作點(diǎn)D坐標(biāo)為（4，0），可得BD＝1，進(jìn)而有SKIPIF1<0，再加上公共角∠PBD＝∠ABP，根據(jù)兩邊對(duì)應(yīng)成比例且?jiàn)A角相等可證△PBD∽△ABP，得SKIPIF1<0等于相似比SKIPIF1<0，進(jìn)而得PD＝SKIPIF1<0AP，所以當(dāng)C、P、D在同一直線上時(shí)，PC+SKIPIF1<0PA＝PC+PD＝CD最小．用兩點(diǎn)間距離公式即求得CD的長(zhǎng)．【詳解】解：（1）直線y＝﹣5x+5，x＝0時(shí)，y＝5∴C（0，5）y＝﹣5x+5＝0時(shí)，解得：x＝1∴A（1，0）∵拋物線y＝x2+bx+c經(jīng)過(guò)A，C兩點(diǎn)∴SKIPIF1<0

解得：SKIPIF1<0∴拋物線解析式為y＝x2﹣6x+5當(dāng)y＝x2﹣6x+5＝0時(shí)，解得：x1＝1，x2＝5∴B（5，0）（2）如圖1，過(guò)點(diǎn)M作MH⊥x軸于點(diǎn)H∵A（1，0），B（5，0），C（0，5）∴AB＝5﹣1＝4，OC＝5∴S△ABC＝SKIPIF1<0AB?OC＝SKIPIF1<0×4×5＝10∵點(diǎn)M為x軸下方拋物線上的點(diǎn)∴設(shè)M（m，m2﹣6m+5）（1＜m＜5）∴MH＝|m2﹣6m+5|＝﹣m2+6m﹣5∴S△ABM＝SKIPIF1<0AB?MH＝SKIPIF1<0×4（﹣m2+6m﹣5）＝﹣2m2+12m﹣10＝﹣2（m﹣3）2+8∴S四邊形AMBC＝S△ABC+S△ABM＝10+[﹣2（m﹣3）2+8]＝﹣2（m﹣3）2+18∴當(dāng)m＝3，即M（3，﹣4）時(shí)，四邊形AMBC面積最大，最大面積等于18（3）如圖2，在x軸上取點(diǎn)D（4，0），連接PD、CD∴BD＝5﹣4＝1∵AB＝4，BP＝2∴SKIPIF1<0∵∠PBD＝∠ABP∴△PBD∽△ABP∴SKIPIF1<0∴PD＝SKIP