初中数学几何题？ 初中数学几何题解题技巧
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以AC的中点（设为D）为圆心，以AC/2为半径作圆。（如图）
因PA^2+PC^2=AC^2, 得角APC=90度。
P必在圆上。
当B、P、D三点一线时，PB有最小值
（DB=PB+PD,PD=AC/2一定， BD为线段时最短，此时PB有最小值）
此时：CD=√3, BC=3 角DCB=90度
得：角DBC为30度，角BDC为60度
所以： PDC是等边三角形。角PCA为60度
PC=√3，PA=3
三角形ACP的面积=（3*√3）/2=1.5√3
解:取AC中点O，连接 OP，BO，
因为，PA²+PC²= AC²,
所以，角APC=90°，
所以，点P在以AC为直径的圆上运动，在△BPO中，BP≥BO-OP，
所以，当点P在线段BO上时，BP有最小值，
因为点0是AC的中点，角APC=90°
所以，PO=AO=CO=√3
因为tan角BOC=CO/BC=√3/3
∴角BOC=60°，
∴△COP是等边三角形，
所以，S△cop=√3/4CO²=√3/4 *3=3√3/4
因为OA=OC
所以△ACP的面积=2S△cop=3√3/2
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绛旓細1锛氳彵褰㈢殑涓ゆ潯瀵硅绾块暱鍒嗗埆鏄1cm鍜2cm,鍒欒繖涓彵褰㈢殑闈㈢Н鏄紙1cm²锛夈2锛氬湪姝ｆ柟褰BCD鎵鍦ㄥ钩闈㈠唴瀛樺湪锛堜腑蹇冩垨閲嶅績锛夛紝浣垮畠涓庡悇瀹氱偣鏋勬垚鐨勬墍鏈変笁瑙掑舰閮戒负绛夎叞涓夎褰3锛氬湪骞宠鍥涜竟褰BCD涓紝DC=2AD锛孧鏄疍C鐨勪腑鐐癸紝鍒欒AMB绛変簬锛90掳锛夈傘愬垎鏋愩戣繃M浣淢N鈭D锛屽垯MADN鍜孧NBC鏄彵褰侫M鍜孊...
绛旓細姝ｆ柟褰㈡婚潰绉細6脳6=36 濡傛灉瀵规鏂瑰舰浠庡乏涓婅椤烘椂閽堢紪鍙凤細涓篘MPQ锛堝乏涓婅涓篘锛屽彸涓婅涓篗锛屽彸涓嬭涓篜锛屽乏涓嬭涓篞锛塎D鍜孮C鐨勪氦鐐逛负O 浣犲彲浠ヨ交鏉捐瘉鏄庝笁瑙掑舰QOP鍜屼笁瑙掑舰MOP鍏ㄧ瓑锛屾墍浠鍒癕P鍜孮P鐨勮窛绂绘槸鐩哥瓑鐨勶紝閭ｄ箞灏辫杩欎釜璺濈涓簒 鎵浠ュ彲浠ヨ交鏉惧緱鍒癉P脳x梅2+MP脳x梅2=MP脳DP梅2 涔熷氨鏄1....
绛旓細(1)銆佲埖ABCD鏄钩琛屽洓杈瑰舰銆傗埓鈭燘=鈭燙DM,AB=CD,BC=AD,鈭礏N=NC.AM=DM,鈭碆N=DM,鍒欌娍ABN鈮屸娍CDM锛(2)銆佺敱AM=MD,BN=NC鍙煡MD=NC,MD鈭C,鈭碝NCD鏄钩琛屽洓杈瑰舰銆傝岀敱AM=NC,AM鈭C鍙煡ANCM涔熸槸骞宠鍥涜竟褰紝鈭碅N鈭C,鈭碘垹AND=90掳,鈭粹垹MOD=90掳鍒橫NCD杩樻槸鑿卞舰銆侻N=NC銆傗垹1=鈭燚NC=鈭2...
绛旓細鍒濅腑鏁板鍑犱綍瀹氱悊闆嗛敠 1銆傚悓瑙掞紙鎴栫瓑瑙掞級鐨勪綑瑙掔浉绛夈3銆傚椤惰鐩哥瓑銆5銆備笁瑙掑舰鐨勪竴涓瑙掔瓑浜庡拰瀹冧笉鐩搁偦鐨勪袱涓唴瑙掍箣鍜屻6銆傚湪鍚屼竴骞抽潰鍐呭瀭鐩翠簬鍚屼竴鏉＄洿绾跨殑涓ゆ潯鐩寸嚎鏄钩琛岀嚎銆7銆傚悓浣嶈鐩哥瓑锛屼袱鐩寸嚎骞宠銆12銆傜瓑鑵颁笁瑙掑舰鐨勯《瑙掑钩鍒嗙嚎銆佸簳杈逛笂鐨勯珮銆佸簳杈逛笂鐨勪腑绾夸簰鐩搁噸鍚堛16銆傜洿瑙掍笁瑙掑舰涓紝鏂滆竟涓...
绛旓細浠C鐨勪腑鐐癸紙璁句负D锛変负鍦嗗績锛屼互AC/2涓哄崐寰勪綔鍦嗐傦紙濡傚浘锛夊洜PA^2+PC^2=AC^2, 寰楄APC=90搴︺侾蹇呭湪鍦嗕笂銆傚綋B銆丳銆丏涓夌偣涓绾挎椂锛孭B鏈夋渶灏忓 锛圖B=PB+PD,PD=AC/2涓瀹氾紝 BD涓虹嚎娈垫椂鏈鐭紝姝ゆ椂PB鏈夋渶灏忓硷級姝ゆ椂锛欳D=鈭3, BC=3 瑙扗CB=90搴 寰楋細瑙扗BC涓30搴︼紝瑙払DC涓60搴 鎵浠...
绛旓細瑙佸浘
绛旓細5.鈭礏D涓烘鏂瑰舰ABCD鐨勫瑙掔嚎锛屸埓AF=CF,鈻矨FD鈮屸柍CFD锛圫AS锛夆垹DAF=鈭燚CF銆傛槗璇佲柍ABE鈮屸柍DCE锛圫AS)鈭粹垹ABE=鈭燚CF=鈭燚AF,鈭.鈭燛AF+鈭燗EB=鈭燗BE+鈭燗EB=90掳鍗矨F鈯E 杩欎簺閮芥槸鍒濅腑棰
绛旓細瑙払=40搴︼紝鎵浠 瑙払AC=90-40=50搴 瑙扖AD=180-50=130搴 瑙扖AE=瑙扗AE=130/2=65搴 鎵浠ヨBAE=瑙払AC+瑙扖AE=50+65=115搴 瑙扙=90-瑙扖AE=25搴
