如图所示，直线AB、CD相交于点O，OE垂直AB，点O为垂足，OF平分角AOC，且角COE＝五分之 如图，直线AB、CD相交于点O，OE⊥AB，点O为垂足，OF...
\u5982\u56fe\u6240\u793a\uff0c\u76f4\u7ebfab\u3001cd\u76f8\u4ea4\u4e8e\u70b9o\uff0coe\u5782\u76f4\u4e8eab\uff0c\u70b9o\u4e3a\u5782\u8db3\uff0cof\u5e73\u5206\u89d2aoc\uff0c\u4e14\u89d2coe=\u4e94\u5206\u8bbe\u2220COE=X\u00b0\uff0c
\u5219\u2220AOC=\u2220AOE+\u2220COE=90\u00b0+X\u00b0\uff0c
\u7531\u9898\u610f\u5f97X=3/5(90+X)
\u89e3\u5f97X=60
\u2234\u2220AOC=150\u00b0\uff0c
\u2235OD\u5e73\u5206\u2220AOC\uff0c
\u2234\u2220COD=1/2\u2220AOC=75\u00b0\uff0c
\u2234\u2220DOF=180\u00b0-\u2220COD=105\u00b0
\u2235OE\u22a5AB\uff0c\u2234\u2220AOE=\u2220BOE=90\u00b0\uff0c\u8bbe\u2220EOC=2x\uff0c\u2220AOC=5x\uff0e\u2235\u2220AOC-\u2220COE=\u2220AOE\uff0c\u22345x-2x=90\u00b0\uff0c\u89e3\u5f97x=30\u00b0\uff0c\u2234\u2220COE=60\u00b0\uff0c\u2220AOC=150\u00b0\uff0e\u2235OF\u5e73\u5206\u2220AOC\uff0c\u2234\u2220AOF=75\u00b0\uff0e\u2235\u2220AOD=\u2220BOC=90\u00b0-\u2220COE=30\u00b0\uff0c\u2234\u2220DOF=\u2220AOD+\u2220AOF=105\u00b0\uff0e
COE=五分之二AOC，OE垂直于AB，所以AOC=90度+五分之二AOC，得出AOC等于150度。AOC+DOA=180度，所以DOA=30度，OF平分AOC，所以AOF=75度，所以DOF=DOA+AOF=30度+75度=105度绛旓細瑙ｏ細鈭鐩寸嚎AB 鈭粹垹AOB锛180 鈭碘垹AOE锛140 鈭粹垹BOE锛濃垹AOB-鈭燗OE锛180-140锛40 鈭礝E骞冲垎鈭燘OD 鈭粹垹DOE锛濃垹BOE锛濃垹BOD/2 鈭粹垹BOD锛2鈭燘OE锛80 鈭电洿绾緼B銆CD鐩镐氦浜鐐筄 鈭粹垹AOC锛濃垹BOD锛80
绛旓細OF涓嶰D鐨勪綅缃叧绯绘槸锛氫簰鐩稿瀭鐩淬傝瘉鏄庯細鍥犱负 鐩寸嚎AB,CD鐩镐氦浜鐐筄锛屾墍浠 瑙扐OB=180搴︼紝鍥犱负 OD骞冲垎瑙払OE锛孫F骞冲垎瑙扐OE锛屾墍浠 瑙扗OE=1/2瑙払OE锛岃EOF=1/2瑙扐OE锛屾墍浠 瑙扗OE+瑙扙OF=1/2瑙払OE+1/2瑙扐OE =1/2(瑙払OE+瑙扐OE)=1/2瑙扐OB =1/2X180搴 =90搴︼紝鎵浠 瑙扗OF=...
绛旓細鈭礝F鏄垹AOE鐨勫钩鍒嗙嚎 鈭燛OF=55掳 鈭粹垹AOE=55掳脳2=110掳 鈭粹垹BOE=180掳锛110掳=70掳 鍙堚埖OD鏄垹BOE鐨勫钩鍒嗙嚎 鈭粹垹BOD=鈭燘OE梅2=70掳梅2=35掳 鈭粹垹AOC=35掳锛堝椤惰鐩哥瓑锛夆埓鈭燙OE=鈭燗OC锛嬧垹AOE=35掳锛110掳=145掳
绛旓細瑙ｏ細璁锯垹AOC锛漍 鈭碘垹AOC锛氣垹AOD锛1锛5, 鈭燗OC锛漍 鈭粹垹AOD锛5X 鈭碘垹AOC+鈭燗OD锛180 鈭碭+5X锛180 X锛30 鈭粹垹AOC锛30, 鈭燗OD锛150 鈭碘垹BOD锛濃垹AOC锛30 锛堝椤惰鐩哥瓑锛夛紝OD骞冲垎鈭燘OE 鈭粹垹DOE锛濃垹BOD锛30 鈭粹垹AOE锛濃垹AOD-鈭燚OE锛150-30锛120 鈭礝F骞冲垎鈭燗OE 鈭粹垹EOF锛濃垹AOE/2锛120/...
绛旓細鍥犱负EO鈯AB 鎵浠モ垹EOA=90掳 鎵浠モ垹AOC=鈭燛OA-鈭燛OC=90掳-35掳=55掳 鈭燗OD=180掳-鈭燗OC=125掳
绛旓細鍥犱负瑙扐OC+瑙払OC=180搴︼紝瑙払OC=20搴︼紝鎵浠ヨAOC=160搴︼紝鍥犱负OE骞冲垎瑙扐OC锛屾墍浠ヨCOE=瑙扐OC/2=160/2=80搴︼紝鍥犱负瑙払OE=瑙扖OE+瑙払OC=80+20=100搴︼紝鎵浠ヨBOE鐨勫害鏁版槸100
绛旓細绛 鈭礝E骞冲垎鈭燘OC 鈭粹垹BOE=鈭燙OE 鈭碘垹BOD:鈭燘OE=2:3 鈭粹垹BOD:鈭燘OE锛氣垹COE=2:3:3 鈭碘垹COD=180掳锛堝钩瑙掓槸180掳锛夆埓鈭燘OD=180掳脳2/8=45掳 鈭粹垹AOC=45掳锛堝椤惰鐩哥瓑锛夆埖OF鈯CD 鈭粹垹AOF=90掳-45掳=45掳 濡傛灉浣犺鍙垜鐨勫洖绛旓紝璇风偣鍑诲乏涓嬭鐨勨滈噰绾充负婊℃剰绛旀鈥濓紝绁濆涔犺繘姝ワ紒
绛旓細鈭礝M鈯AB锛鈭1=鈭2锛屸埓鈭1+鈭燗OC=鈭2+鈭燗OC=90掳锛屽嵆鈭燙ON=90掳锛涘張鈭燦OC+鈭燦OD=180掳锛屸埓鈭燦OD=90掳锛
绛旓細璁锯垹COE=X掳锛屽垯鈭燗OC=鈭燗OE+鈭燙OE=90掳+X掳锛岀敱棰樻剰寰梄=3/5(90+X)瑙ｅ緱X=60 鈭粹垹AOC=150掳锛屸埖OD骞冲垎鈭燗OC锛屸埓鈭燙OD=1/2鈭燗OC=75掳锛屸埓鈭燚OF=180掳-鈭燙OD=105掳
绛旓細鈭燼oc+鈭燼od=180掳锛屾墍浠モ垹aoc=45掳锛屸垹aod=135掳 瀵归《瑙掔浉绛夛紝鎵浠モ垹bod=鈭燼oc=45掳锛屸垹boc=鈭燼od=135掳
