AD为三角形ABC的外接圆圆O的直径，AD垂直于BC 垂足为F 角ABC的平分线角AD于点E 连接 如图，AD为△ABC外接圆的直径，AD⊥BC，垂足为点F，∠...
\u5982\u56fe AD\u4e3a\u4e09\u89d2\u5f62\u5916\u63a5\u5706\u7684\u76f4\u5f84 AD\u4e04BC \u5782\u8db3\u4e3a\u70b9F \u89d2ABC\u7684\u5e73\u5206\u7ebf\u4ea4AD\u4e8e\u70b9E1\u3001\u2235AD\u662f\u76f4\u5f84,\u5219\u5fc5\u8fc7\u5706\u5fc3O,AD\u22a5BC,
\u5373OF\u22a5BC,
\u2234F\u662fBC\u7684\u4e2d\u70b9,\uff08\u5f26\u5fc3\u8ddd\u5782\u76f4\u5e73\u5206\u5f26\uff09,
\u2234AD\u662fBC\u7684\u5782\u76f4\u5e73\u5206\u7ebf,
\u2235D\u2208AD,
\u2234BD=CD.
2\u3001\u2235AD\u662fBC\u7684\u5782\u76f4\u5e73\u5206\u7ebf,
\u2234AB=AC,
\u2234\u25b3ABC\u662f\u7b49\u8170\u25b3,
\u2234AF\u662f\u3008BAC\u5e73\u5206\u7ebf,
\u2235BE\u662f\u3008ABC\u5e73\u5206\u7ebf,
\u2234E\u662f\u5185\u5fc3,\uff08\u4e09\u89d2\u5f62\u4e09\u6761\u89d2\u5e73\u5206\u7ebf\u7684\u4ea4\u70b9\uff09
\u2234CE\u662f\u3008C\u5e73\u5206\u7ebf,
\u3008BEF=\u3008BAE+\u3008ABE=\u3008A/2+\u3008B/2,
\u3008EBD=\u3008B/2+\u3008FBD,
\u3008FBD=\u3008DAC=\u3008A/2,\uff08\u540c\u5f27\u5706\u5468\u89d2\u76f8\u7b49\uff09,
\u3008FBD=\u3008B/2+\u3008A/2,
\u2234\u3008DBE=\u3008DEB,
\u2234BD=DE,
\u540c\u7406\u53ef\u8bc1DE=DC,
BD=DE=DC,
\u2234B\u3001E\u3001C\u4e09\u70b9\u5728\u4ee5D\u4e3a\u5706\u5fc3,\u4ee5DB\u4e3a\u534a\u5f84\u7684\u5706\u4e0a.
\u2220EBF\u2260\u2220BAD,\u9664\u975e\u662f\u6b63\u4e09\u89d2\u5f62.
\u89e3\uff1a\uff081\uff09\u2235 \u4e3a\u76f4\u5f84\uff0c \u2234 \u2234 \u3002\uff082\uff09\u7b54\uff1aB\uff0cC\uff0cE\u4e09\u70b9\u5728\u4ee5D\u4e3a\u5706\u5fc3\uff0c\u4ee5 \u4e3a\u534a\u5f84\u7684\u5706\u4e0a\u7406\u7531\uff1a\u7531\uff081\uff09\u77e5 \u2234 \u2235 \uff0c \u2234 \u2234 \u7531\uff081\uff09\u77e5 \u2234 \u2234B\uff0cC\uff0cE\u4e09\u70b9\u5728\u4ee5D\u4e3a\u5706\u5fc3\uff0c\u4ee5 \u4e3a\u534a\u5f84\u7684\u5706\u4e0a\u3002
你。
绛旓細璇佹槑锛氬洜涓篈D鏄涓夎褰ABC鐨勫鎺ュ渾鐨勭洿寰 鎵浠ヨABD=90搴 AD=2AO 鍥犱负AE鍨傜洿BC 鎵浠ヨAEC=90搴 鍥犱负瑙扗=瑙扖=寮B/2 鎵浠ョ洿瑙掍笁瑙掑舰ABD鍜岀洿瑙掍笁瑙掑舰AEC鐩镐技 鎵浠B/AE=AD/AC 鍗矨B涔樹互AC=AE涔樹互AD 鎵浠B涔樹互AC=2AE涔樹互AO
绛旓細鍥犱负瑙払鍦ㄥ渾涓婂鐨勫姬涓篈C锛屼笌瑙扐DC鎵瀵瑰姬闀夸竴鑷达紝鎵浠ヨB=瑙扐DC 鍦ㄧ洿瑙涓夎褰ADC涓紝AD=3锛孉C=2锛屾墍浠D=鏍瑰彿5 cosB = cos(ADC) = (鏍瑰彿5) / 3
绛旓細鈭粹垹ACD鏄洿瑙掞紙鍗婂渾涓婄殑鍦嗗懆瑙掓槸鐩磋锛夆埓cos鈭燚=CD锛AD 鈭2/3=CD锛氾紙3/2脳2锛夎В鍑恒CD=2 浜庢槸锛欰C=鏍瑰彿涓3骞虫柟锛2骞虫柟锛濇牴鍙5
绛旓細涓よ鐩哥瓑,寮B=寮C,鎵浠,AB=AC,鎵浠,瑙 ABC=ACB 瑙扐CB=ADB DE骞宠BC,ABC=瑙扙 鎵浠 瑙扐DB=瑙扙 鎵浠鐐逛綅缃拰瑙扙澶у鏃犲叧,褰揇浣嶄簬寮C涓偣鏃,AD鏄鐩村緞,ED鏄垏绾,杩炴帴BD,璁剧洿寰剎 sinBAD=3/5=鈭(x^2-5^2)/x 鎵浠=25/4 鍗婂緞 25/8 鍙傝冭祫鏂欙細http://zhidao.baidu.com/question/1...
绛旓細鍥犱负鈭燛=鈭燘锛岋紙鍚屽姬鎵瀵圭殑鍦嗗懆瑙掞級.鈭燘=鈭燙AD锛屾墍浠モ垹CAD+鈭燛AC=90掳锛屽嵆EA鍨傜洿浜嶢D锛孉E鏄渾o鐨勭洿寰勶紝鎵浠AD鏄渾o鐨勫垏绾裤 2锛夛紝浣溾棆o鐨勭洿寰凙E杩炵粨EC锛屽湪鈻矰AC鍜屸柍DAB涓 鈭燚鍏敤锛孉D²=DCBD锛屽嵆AD/BD=DC/AD锛屾墍浠モ柍DAC鈭解柍DBA锛屸垹DAC=鈭燘銆備豢1锛夊彲璇丄D鏄渾o鐨勫垏绾裤
绛旓細杩炵粨CD 鍥犱负鈭燘鍜屸垹ADC瀵瑰簲浜庡悓涓鏉″渾寮C 鎵浠モ垹B=鈭燗DC 鍥犱负AD鏄鐩村緞锛屾墍浠モ垹ACD=90掳 鐢遍鐭D=3锛孉C=2 鍒檚inB=sin鈭燗DC=AC/AD=2/3
绛旓細瑙ｏ細锛1锛夎瘉鏄庯細寤堕暱DE浜も姍0浜嶩锛岃繛鎺H銆丅H锛庡垯鍥涜竟褰HEF涓虹煩褰紝鈭碅F=EH锛孉H鈭F锛屸埓鈭燞AB=鈭ABC锛屸埓BH=AC锛屸埓Rt鈻矪EH鈮孯t鈻矯FA锛屸埓BE=CF锛涳紙2锛夎繛鎺D锛岃繛鎺O骞跺欢闀夸氦DE浜嶱鐐癸紟鍒欌柍AFO鈮屸柍DPO锛屸埓AF=DP锛孫F=OP锛屸埓OG=1/2PE锛屸埓PE=2锛屸埓AF=DP=1鈭礑E=BF=CE锛屸埓鈭燘CD=45掳...
绛旓細璇佹槑锛氣埖AE鏄渾O鐨勭洿寰 鈭粹垹ABE锛90 鈭粹垹BAE+鈭燛锛90 鈭碘垹E銆佲垹C鎵瀵瑰簲鍦嗗姬閮戒负鍔ｅ姬AB 鈭粹垹E锛濃垹C 鈭粹垹BAE+鈭燙锛90 鈭AD鈯C 鈭粹垹CAD+鈭燙锛90 鈭粹垹BAE锛濃垹CAD
绛旓細浣犮
绛旓細璇佹槑锛氳繛鎺E锛 鈭礎E鏄姍O鐨勭洿寰勶紝 鈭粹垹ABE=90掳锛 鈭粹垹BAE+鈭燛=90掳锛 鈭AD鏄鈻ABC杈逛笂鐨勯珮锛 鈭粹垹ADC=90掳锛 鈭粹垹CAD+鈭燗CB=90掳锛 鈭碘垹E=鈭燗CB锛 鈭粹垹BAE=鈭燙AD锛
