2012年2月觀周刊第8期總568期.February 2012 M i有關(guān)調(diào)和性質(zhì)的應(yīng)用°戶(hù)淑茹②趙臨龍(安康學(xué)院數(shù)學(xué)與應(yīng)用數(shù)學(xué)研究所安康725000中圖分類(lèi)號(hào):0185.1文獻(xiàn)標(biāo)識(shí)碼:A文章編號(hào):1008-925X(2012)02-0229-01摘要:利用調(diào)和分割的定義解決初等幾何問(wèn)題:如何做調(diào)和共軛點(diǎn)、證明一角的兩邊和它的內(nèi)外交平分線(xiàn)成調(diào)和線(xiàn)束、一線(xiàn)段被它的中點(diǎn)和這直線(xiàn)上的無(wú)穿遠(yuǎn)點(diǎn)所調(diào)和分割、以及如何用完全四點(diǎn)形的調(diào)和性質(zhì)作圖關(guān)鍵詞:調(diào)和分割中點(diǎn)無(wú)窮遠(yuǎn)點(diǎn)完全四點(diǎn)形The application concerning concordance propertyHu Shuru Zhao LinlongAbstract: The definition made use of to temper partition solution the elementary grade is several what problem: How temper conju-gate point and prove the both sides of dime with its inside the diplomacy divide equally line to become to temper line to tie, front-linesegment drive its middle point and this straight line up of the everlasting far order in harmony with partition, and how use up the inharmony with property of whole squareses to make diagram.Keywords: In harmony with partition MiddleThe everlasting far orders Complete square調(diào)和分割的定義:設(shè)A,B,C,D為共線(xiàn)的四點(diǎn),我們把由于∠2=∠TOB,∠1=∠BOTAG:B犯定義為這四點(diǎn)按著順序的交比,并以符號(hào)(ABCD)∴∠1=∠2,同理:∠3=∠4表達(dá)CAB, CD)=B當(dāng)(AB,CD)=-1時(shí),即C點(diǎn)分割線(xiàn)(由三角形中一個(gè)角的內(nèi)角和外交平分線(xiàn)和對(duì)邊的交點(diǎn),調(diào)和分割對(duì)點(diǎn)段AB的值和D點(diǎn)分割線(xiàn)段AB的值只差一個(gè)符號(hào)(因而一個(gè)是從中學(xué)幾何,我們知道三角形中一個(gè)角的內(nèi)角和外角平分線(xiàn)內(nèi)分點(diǎn),一個(gè)是外分點(diǎn)),這時(shí)說(shuō)C,D兩點(diǎn)調(diào)和分割線(xiàn)段AB或?qū)?duì)邊分成兩線(xiàn)段的比值都和兩鄰邊成比例(注意中學(xué)幾何講的對(duì)C與D于線(xiàn)段AB成調(diào)和共軛點(diǎn)偶是絕對(duì)量!),可見(jiàn)兩條平分角線(xiàn)和對(duì)邊的交點(diǎn),調(diào)和分割對(duì)邊但是,在調(diào)和分割中,兩對(duì)點(diǎn)的關(guān)系是完全對(duì)等的,即當(dāng)C與例3一線(xiàn)段被它的中點(diǎn)和這直線(xiàn)上的無(wú)窮遠(yuǎn)點(diǎn)所調(diào)和D調(diào)和分割線(xiàn)段AB時(shí),A與B也調(diào)和分割CD.若(AB,CD)=分割。1,便也有(CD,AB)=-1。證明:設(shè)C點(diǎn)為線(xiàn)段AB的中點(diǎn),D。為直線(xiàn)AB的無(wú)窮遠(yuǎn)點(diǎn)例1試證一角的兩邊和它的內(nèi)外交平分線(xiàn)成調(diào)和線(xiàn)束。證明:以下引用的角均系有向角,即則(AB,CD)=xABC(b,c),(b,d)=90-(c,b由于(ABD)=mD=1,因此(AB,CD=)=(ABC)=(a,a)=90°+(a,c)=90°-(b,c)由上題還可知,共線(xiàn)四點(diǎn)的交比,若第四點(diǎn)為無(wú)窮遠(yuǎn)點(diǎn),那么(ab, cd)=sin(a, c)) sin(b,d)- sin[ 90"-(c,b)]一交比值等于前三點(diǎn)的簡(jiǎn)比值sin(a,d)·sin(b,c)sin[90°+(a,c)]例4調(diào)和點(diǎn)列作圖:已知A,B,C求第四點(diǎn)Dcos(c, b)可以利用完全四點(diǎn)形的調(diào)和性質(zhì),解決問(wèn)題cos(a所以a,b調(diào)和分割c,d例2如圖1所示,AB為圓之直徑,C為直徑延長(zhǎng)線(xiàn)上一點(diǎn)從C向圓引切線(xiàn)CT,證明:若C在線(xiàn)段AB本身上,如何作它的調(diào)和共軛點(diǎn)?完全四點(diǎn)形:如圖3,平面內(nèi)無(wú)三點(diǎn)共線(xiàn)的四點(diǎn)及其兩兩聯(lián)線(xiàn)所D構(gòu)成的圖形稱(chēng)為完全四點(diǎn)形.并且完全四點(diǎn)形通過(guò)每一個(gè)對(duì)角點(diǎn)有組調(diào)和線(xiàn)束,即通過(guò)這對(duì)對(duì)角點(diǎn)的兩邊和對(duì)角三角形的兩條邊?，F(xiàn)在給出第四點(diǎn)D作法:如圖4.①過(guò)A,B,C作線(xiàn)外一點(diǎn)S,作線(xiàn)束SA,SB,SC作圖:如圖2,過(guò)點(diǎn)C作TT⊥AB,直接作TD⊥OT交AB延②在SC上取一點(diǎn)Q,并 AQXSB=T, BQX SA=R長(zhǎng)線(xiàn)于點(diǎn)D,即D調(diào)和分割定點(diǎn)A,B,C③作RT×ABC于點(diǎn)D,即D點(diǎn)為(AB,CD)=-1的第證明:需證T在AB上的垂直射影,D是C對(duì)于AB的調(diào)和四點(diǎn)共軛點(diǎn)需證:A.Bb=-(Ab·B參考文[1]朱德樣,朱維宗[2]李文銘,初等幾H中國(guó)煤化工樓CNMHG①資助課題:映西省特色專(zhuān)業(yè)建設(shè)項(xiàng)目(2011-59);安康學(xué)院大學(xué)生科技創(chuàng)新項(xiàng)目(2011 AKXYDXS09)Q戶(hù)淑茹(1989-),女,陜西省商洛市人,現(xiàn)為安康學(xué)院數(shù)學(xué)系09級(jí)學(xué)生。229