△ABC中，AD是BC边上的高线，BE是一条角平分线，它们相交于点P已知∠EPD=125°求∠BAD
\u5982\u56fe\uff0c\u5728\u4e09\u89d2\u5f62ABC\u4e2d\uff0c\u5df2\u77e5\u89d2ABC=30\u5ea6\uff0c\u70b9D\u5728BC\u4e0a\uff0c\u70b9E\u5728AC\u4e0a\uff0c\u89d2BAD=\u89d2EBC,AD\u4ea4BE\u4e8eF (1) \u6c42\u89d2BFD\u7684\u5ea6\u6570\u5206\u6790\uff1a\uff081\uff09\u5148\u6839\u636e\u2220ABC=30\u00b0\uff0c\u2220BAD=\u2220EBC\u53ef\u77e5\uff0c\u2220BAD+\u2220ABD=\u2220EBC+\u2220ABD=\u2220ABC=30\u00b0\uff0c\u518d\u6839\u636e\u4e09\u89d2\u5f62\u5916\u89d2\u7684\u6027\u8d28\u5373\u53ef\u5f97\u51fa\u7ed3\u8bba\uff1b
\uff082\uff09\u5148\u6839\u636eEG\u2225AD\uff0c\u2220BFD=30\u00b0\u53ef\u77e5\u2220BEG=30\u00b0\uff0c\u518d\u6839\u636eEH\u22a5BE\u53ef\u77e5\u2220BEH=90\u00b0\uff0c\u6545\u53ef\u6c42\u51fa\u2220HEG\u7684\u5ea6\u6570\uff0e
\u89e3\u7b54\uff1a\u89e3\uff1a\uff081\uff09\u2235\u2220ABC=30\u00b0\uff0c\u2220BAD=\u2220EBC\uff0c
\u2234\u2220BAD+\u2220ABD=\u2220EBC+\u2220ABD=\u2220ABC=30\u00b0\uff0c
\u2235\u2220BFD\u662f\u25b3ABF\u7684\u5916\u89d2\uff0c
\u2234\u2220BFD=\u2220BAD+\u2220ABD=30\u00b0\uff1b
\uff082\uff09\u2235EG\u2225AD\uff0c\u2220BFD=30\u00b0\uff0c
\u2234\u2220BEG=\u2220BFD=30\u00b0\uff0c
\u2235EH\u22a5BE\uff0c
\u2234\u2220BEH=90\u00b0\uff0c
\u2234\u2220HEG=\u2220BEH-\u2220BDG=90\u00b0-30\u00b0=60\u00b0\uff0e
\u8fd8\u662f\u6211\u6765\u5e2e\u697c\u4e3b\u56de\u7b54\u8fd9\u9053\u9898\u76ee\u5427\uff01\u89c9\u5f97\u6709\u7528\u4e00\u5b9a\u8981\u91c7\u7eb3\u554a\uff01
∠EPD=125°,则∠BPD=55°,∠PBD=90°-55°=25°=∠ABP,则
∠BAD=∠BPD-∠ABP=55°-25°=30°如果还有不明白的，可以再问，望采纳！
绛旓細鍦ㄢ柍ABD涓,鐢变簬鈭燘=30掳,BD=10,鍒欙細AD=BD/(鈭3)=(10鈭3)/3 鍦鈻矨DC涓,AC=鈭2AD=(鈭2)脳[(10鈭3)/3]=(10鈭6)/3
绛旓細鈥滄暟鐞嗙瓟鐤戝洟鈥濅负鎮ㄨВ绛旓紝甯屾湜瀵逛綘鏈夋墍甯姪銆傝瘉鏄庯細寤堕暱CB鍒癎浣緽G=BE锛岃繛鎺G锛孉D鏄疊C杈逛笂鐨勯珮,CE鏄疉B杈逛笂鐨勪腑绾,鍒欑洿瑙掍笁瑙掑舰ADB涓瑼E=BE=DE锛屽張DC=AE,鍒橝E=BE=DE=DC锛屽垯鈭燘DE=鈭燚BE锛岃屸垹BDE+鈭燙DE=鈭燚BE+鈭燛BG=180掳锛屽洜姝も垹CDE=鈭燛BG锛屾晠锛鈻CDE鈮屸柍EBG锛屽垯 鈭燚CE=鈭燘EG锛孍C=EG...
绛旓細杩囩偣E浣淓F鈯BC浜嶧,鍒欐湁锛欵F鈥AD .鍥犱负,EF = BE路sin鈭燙BE = (1/2)BE ,鎵浠,BE = 2EF = AD .鍥犱负,EF鈥朅D ,AE = EC ,鎵浠,EF鏄鈻ACD鐨勪腑浣嶇嚎,鍙緱锛欰D = 2EF = BE .
绛旓細鍒嗘瀽锛氭牴鎹笁瑙掑舰鐨勫唴瑙掑拰瀹氱悊鍙婅骞冲垎绾跨殑鎬ц川姹傝В锛庤В锛氣埖鍦鈻矨BC涓紝AE鏄垹BAC鐨勫钩鍒嗙嚎锛屼笖鈭燘=40掳锛屸垹C=60掳锛屸埓鈭燘AE=鈭燛AC= 1/2锛180掳-鈭燘-鈭燙锛= 1/2锛180掳-40掳-60掳锛=40掳锛庡湪鈻矨CD涓紝鈭燗DC=90掳锛屸垹C=60掳锛屸埓鈭燚AC=180掳-90掳-60掳=30掳锛屸垹EAD=鈭燛AC-鈭燚AC...
绛旓細涓夎褰abc涓紝鈭燽=75搴,鈭燾=45搴︼紝鍒欌垹bac=60搴 ae骞冲垎鈭燽ac锛屽垯鈭燽ae=鈭爀ac=30搴 ad鏄痓c涓婄殑楂橈紝ad鍨傜洿bc锛屸垹adb=鈭燼dc=90搴 涓夎褰ac涓紝鈭燾=45搴︼紝鈭燼dc=90搴︼紝鍒欌垹dac=45搴︼紝鍙堚垹eac=30搴 鍒欌垹dae=15搴 鈭燼ec=鈭燿ae+鈭燼de=15+90=105搴 ...
绛旓細AD=3cm,EC=2cm. 璇曢鍒嗘瀽锛氫笁瑙掑舰鐨勯潰绉叕寮忕瓑浜庡簳涔樹互楂樼殑涓鍗,涓夎褰竴杈圭殑涓嚎骞冲垎杩欐潯杈,鐢遍,鍦鈻矨BC涓紝AD鏄疊C杈逛笂鐨勯珮锛孲 鈻矨BC = BC脳AD,鍥犱负BC="4cm," S 鈻矨BC =6cm 2 ,鎵浠D=3cm,鍥犱负AE鏄疊C涓婄殑涓嚎,鎵浠ョ偣E鏄竟BC鐨勪腑鐐,鎵浠E=EC= BC=2cm.璇曢瑙ｆ瀽锛氬湪鈻矨BC涓...
绛旓細璇佹槑锛1銆佲埖AD鈯BC 鈭碅B²-BD²锛滱D²锛孉C²-CD²锛滱D²鈭碅B²-BD²锛滱C²-CD²鈭碅B²-AC²锛滲D²-CD²2銆佲埖BC锛滲D+CD 鈭碆D²-CD²锛濓紙BD+CD锛壝楋紙BD-CD锛夛紳BC脳锛圔D-CD锛夆埓AB²-AC&...
绛旓細鈭AD=BD 鈭磋鈭燘AD=鈭燚BA=x掳锛屸埖AB=AC=CD 鈭粹垹CAD=鈭燙DA=鈭燘AD+鈭燚BA=2x掳锛屸垹DBA=鈭燙=x掳锛屸埓鈭燘AC=3鈭燚BA=3x掳锛屸埖鈭ABC+鈭燘AC+鈭燙=180掳 鈭5x=180掳锛屸埓鈭燚BA=36掳 鈭粹垹BAC=3鈭燚BA=108掳锛
绛旓細鈭AD鏄疊C杈逛笂鐨勯珮锛鈭燛AD=5掳锛屸埓鈭燗ED=85掳锛屸埖鈭燘=50掳锛屸埓鈭燘AE=鈭燗ED-鈭燘=85掳-50掳=35掳锛屸埖AE鏄垹BAC鐨勮骞冲垎绾匡紝鈭粹垹BAC=2鈭燘AE=70掳锛屸埓鈭燙=180掳-鈭燘-鈭燘AC=180掳-50掳-70掳=60掳锛
绛旓細璇佹槑锛氬洜涓篈B=AC,AD鏄疊C杈逛笂鐨勯珮 鎵浠D鍨傜洿骞冲垎BC锛屽嵆BD=DC=1/2BC 涓擜D骞冲垎瑙払AC 鍥犱负瑙払AC=120掳 鎵浠モ垹BAD=鈭燚AC=1/2鈭燘AC=60掳 鏍规嵁浠ヤ笂鏉′欢鍙緱鈭燗BD=鈭燘DA-鈭燘AD=90掳-60掳=30掳 鍙堝洜涓篋E鈯B 鎵浠E=1/2BD 鍚岀悊鍙瘉DF=1/2DC 鎵浠E+DE=1/2BD+1/2DC=1/2(BD+DC)=1...
