如图，在rt三角形abc中，角b等于90度，ab等于 6，角c等于30度，点d从点c出发沿ca方向 在RT三角形ABC中，角B=90度，AC=10，角C=30度...
\u5982\u56fe1\uff0c\u5728\u4e09\u89d2\u5f62abc\u4e2d\uff0c\u89d2b\uff1d90\u5ea6\uff0c\u89d2c\uff1d30\u5ea6\u3002\u70b9d\u4ece\u70b9c\u51fa\u53d1\u6cbfca\u4ee52\u4e2a\u5355\u4f4d\u2571s\u7684\u901f\u5ea6\u5411\u7ec8\u4e00\u95ee\u4e0d\u7528\u8bf4\u4e86\u5427\uff0c\u697c\u4e3b\uff0c\u4e8c\u5927\u9898\u4e00\u5c0f\u95ee\uff0ct\u7b49\u4e8e\u4e09\u5206\u4e4b\u4e94\u500d\u6839\u53f7\u4e09\uff0c\u4e8c\u95ee\u56db\u5206\u4e4b\u4e94\u500d\u6839\u53f7\u4e09
\u770b\u5230\u8fd9\u65e5\u671f\u5c31\u4e0d\u60f3\u56de\u7b54
\u5982\u56fe,\u56db\u8fb9\u5f62ABCD\u4e2d,AC=6,BD=8,\u4e14AC\u22a5BD\u987a\u6b21\u8fde\u63a5(1)\u8bc1\u660e\uff1aCD=2T\u6545DF=2T*1/2=T
\u800cAE=T\uff0c\u6545AE=DF
\u53c8AE\u2225DF\uff0c
\u6545\u56db\u8fb9\u5f62AEDF\u662f\u5e73\u884c\u56db\u8fb9\u5f62\uff08\u4e00\u7ec4\u5bf9\u8fb9\u5e73\u884c\u4e14\u76f8\u7b49\u7684\u56db\u8fb9\u5f62\u662f\u5e73\u884c\u56db\u8fb9\u5f62\uff09\u3002
\uff082\uff09BC=5\u221a3\uff0cAB=5\u221a3/\u221a3=5\uff0cAC=2AB=10
\u53ea\u9700AE=AD=AC-CD\uff0c\u4e5f\u5373
T=10-2T
\u5f97T=10/3
(3)EF\u5e73\u5206\u25b3ABC\u7684\u9762\u79ef\uff0c\u4e5f\u53732SRT\u25b3BEF=SRT\u25b3ABC
BE*BF=1/2*AB*BC
(5-T)(5\u221a3-\u221a3T)=1/2*5*5\u221a3
\u89e3\u5f97
T=5-5\u221a2/2 \uff08T=5+5\u221a2/2>5\u820d\u53bb \uff09
\uff084\uff09\u5f53\u4e14\u4ec5\u5f53<EDF=90\u00b0\u65f6\uff0c\u25b3DEF\u223d\u25b3ABC\u3002
\u6b64\u65f6AE+DF=AB
T+1/2*2T=5
\u89e3\u5f97T=5/2
(5)\u5f53\u4e14\u4ec5\u5f53<EDF=90\u00b0\u65f6\uff0c\u56db\u4e2a\u4e09\u89d2\u5f62CDF,ADE,DEF,ABC\u90fd\u76f8\u4f3c\u3002
\u6545T=5/2
CD=2t AE=t
(1)DF:AB=CD:AC DF:5=2t:10 DF=t=AE
(2)假设能，则AD=10-2t=t=AE t=10/3
此时CD：AC=2t:10=2:3 =CF：CB BF:BC=1:3 BE:BA=1-AE:AB=1-2/3=1/3
所以EF//AC
所以为菱形成立
(3)
CF:CB=CD:CA=t:5
三角形DEF为直角三角形，只能是∠EDF=90°，则四边形BEDF为矩形
DE=BF,DF=BE
因为DF=AE
所以此时E为AB的中点，所以t=5/2/1=2.5
绛旓細鍦≧T涓夎褰BC涓,瑙C=90搴,BD鏄瑙扐BC鐨勫钩鍒嗙嚎浜C浜嶥,鑻D=3,鈻矨BC鐨勯潰绉负12,鍒橝B= AB+BC=8--- (1)CD/AD=BC/AB寰桟D/AC=BC/(AB+BC)=BC/8 CD=3鍗矨C*BC=24---(2)AB^2-BC^2=AC^2 寰8锛圓B-BC)=AC^2---(3)瑙ｏ紙1锛夈侊紙2锛夈侊紙3锛夌殑鏂圭▼缁勫緱 AB= ...
绛旓細鍥犱负CD鏄骞冲垎绾 鎵浠ヨDCF涓45搴 鍦ㄧ洿瑙掍笁瑙掑舰CDF涓璂F=CF 鎵浠ュ洓杈瑰舰DECF鏄鏂瑰舰 鍏崇郴 锛氾紙1锛変笁瑙掑舰涓夊唴瑙掑拰绛変簬180掳銆傦紙2锛変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傦紙3锛変笁瑙掑舰鐨勪竴涓瑙掑ぇ浜庝换浣曚竴涓拰瀹冧笉鐩搁偦鐨勫唴瑙掋傦紙4锛変笁瑙掑舰涓よ竟涔嬪拰澶т簬绗笁杈癸紝涓よ竟涔嬪樊灏忎簬绗笁杈广傦紙...
绛旓細濡傚浘,鍦≧t涓夎褰BC 涓,瑙C=90搴,D涓篈C涓婁竴鐐,CD=3,AD=BD=5,姹傝A鐨勪笁涓 濡傚浘,鍦≧t涓夎褰BC涓,瑙C=90搴,D涓篈C涓婁竴鐐,CD=3,AD=BD=5,姹傝A鐨勪笁涓笁瑙掑嚱鏁板... 濡傚浘,鍦≧t涓夎褰BC涓,瑙扖=90搴,D涓篈C涓婁竴鐐,CD=3,AD=BD=5,姹傝A鐨勪笁涓笁瑙掑嚱鏁板 灞曞紑 鎴戞潵绛 1涓洖绛 #鐑...
绛旓細RT鈯緼BC缁曠偣C椤烘椂閽堟棆杞緱RT鈯緿EC 鈭碅B涓嶥E鏄搴旇竟 鈭碅B涔熸槸鏂滆竟 鈭鍦≧T鈯ABC涓 鈭燗=30掳鈭粹垹B=60掳 鍙堚埖RT鈯緼BC缁曠偣C椤烘椂閽堟棆杞緱RT鈯緿EC 鈭碈D=CB =2 鈭粹娍CDB鏄瓑杈涓夎褰鍗斥垹BCD=60掳 鈭存棆杞n=60掳 璁綝E涓嶢C浜や簬鐐笷 鍒橲闃村奖=S涓夎褰DF 鈭碘垹BCD=60掳鈭...
绛旓細杩嘍鐐逛綔BC杈逛笂鐨勫瀭绾匡紝鍒橠E鈯C,鍙互璇佹槑鍑篟t鈻睞BD鈮孯t鈻睧CD 鍏ㄧ瓑浜涓夎褰BDE涓変釜涓夎褰㈤兘鍙互璇佹槑鍏ㄧ瓑 鎺ョ潃璇佹槑涓夎褰DE鍏ㄧ瓑浜庝笁瑙掑舰CDE.鍒橞E=CE=1/2 BC=5 鎵浠D=5²-3²=4.璇佹槑涓変釜涓夎褰㈠叏绛
绛旓細绛旓細鍥犱负锛歋鈻矨EG=(S鍥涜竟褰BCG)/3=(S鈻ABC-S鈻矨EG)/3 鎵浠ワ細S鈻矨EG锛歋鈻矨BC=1:4鈥︹︼紙1锛夊洜涓猴細EF//BC 鎵浠ワ細RT鈻矨EG鈭絉T鈻矨BC鈥︹︼紙2锛夌敱锛1锛夊拰锛2锛夊緱锛欰G/AC=EG/BC=1/2 鎵浠ワ細EF鏄腑浣嶇嚎銆傛墍浠ワ細鐐笷鏃禦T鈻矨CD鏂滆竟AD涓婄殑涓偣銆傛墍浠ワ細CF=AD/2 鎵浠ワ細CF/AD=1/2 ...
绛旓細鎵浠ヨAED=瑙払ED=90搴 鍥犱负AD骞冲垎瑙払AC 鎵浠ヨCAD=瑙払AD 鍥犱负瑙扖=90搴 鎵浠ヨC=瑙扐ED=90搴 鍥犱负AD=AD 鎵浠ヤ笁瑙掑舰ACD鍏ㄧ瓑涓夎褰ED (AAS)鎵浠C=AE CD=ED 鍥犱负AC=BC 鎵浠E=BC 涓夎褰BC鏄瓑鑵鐩磋涓夎褰 鎵浠ヨB=45搴 鍥犱负瑙払+瑙払ED+瑙払DE=180搴 鎵浠ヨB=瑙払ED=45搴 鎵浠D=EB 鎵...
绛旓細瑙ｏ細浣淧D鈯C浜嶥锛孭E鈯C浜嶦锛濡傚浘锛AP= 鏍瑰彿2t锛孊Q=tcm锛岋紙0鈮锛6锛夆埖鈭燙=90掳锛孉C=BC=6cm锛屸埓鈻ABC涓鐩磋涓夎褰紝鈭粹垹A=鈭燘=45掳锛屸埓鈻矨PE鍜屸柍PBD涓虹瓑鑵扮洿瑙掍笁瑙掑舰锛屸埓PE=AE=鏍瑰彿2/2 AP=tcm锛孊D=PD锛屸埓CE=AC-AE=锛6-t锛塩m锛屸埖鍥涜竟褰ECD涓虹煩褰紝鈭碢D=EC=锛6-t锛塩m锛...
绛旓細锛1锛夎瘉鏄庯細鍥犱负 鍦≧T涓夎褰BC涓,瑙ACB=90搴,CD鍨傜洿浜嶢B,鎵浠 涓夎褰CD鐩镐技浜庝笁瑙掑舰CBD,鎵浠 瑙扐CD=瑙扖BD,瑙扐=瑙払CM,鍚岀悊鍥犱负 CD鍨傜洿浜嶢B,DH鍨傜洿浜嶣M,鎵浠 涓夎褰DH鐩镐技浜庝笁瑙掑舰DBH,鎵浠 瑙扢DH=瑙扗BH,鍥犱负 瑙扐CD=瑙扢DH+瑙扙,瑙扖BD=瑙扗BH+瑙扖BM,鎵浠 瑙扙=瑙扖BM,鍙堝洜涓 瑙扐=瑙払CM,...
绛旓細BC=16鈭3锛孉E=CE=4鈭6锛屾槗绠楀嚭BD=AD=DC=8鈭3锛孋G=EG=4鈭3锛孌G=CD/2=4鈭3锛孊GBD+DG=12鈭3 BE²=AB²+AE²=480 鏄撹瘉Rt鈻矪GE鈭絉t鈻矪EF BF/BE=BE/BG BF=BE²/BG=(40鈭3)/3 CF=BC-BF=16鈭3-(40鈭3)/3=(8鈭3)/3 鈭碨鈻矯FE=1/2脳CF脳EG=1/2...
