tag:blogger.com,1999:blog-6933544261975483399.post6400596938738292007..comments2023-03-25T01:08:45.796-07:00Comments on Go Geometry (Problem Solutions): Problem 339. Triangle, Angle Bisectors, Perpendiculars, Interior Point, DistancesAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger4125tag:blogger.com,1999:blog-6933544261975483399.post-29749210748210001562015-12-25T19:41:34.497-08:002015-12-25T19:41:34.497-08:00http://s2.postimg.org/hpwg8xnpl/pro_339.png1. Cal...http://s2.postimg.org/hpwg8xnpl/pro_339.png1.<br /> Call Points M1, H1 as the projections of point E over segments AC and BC<br />Call Points M2, G1 as the projections of point D over segments AC and AB<br />Let x=EF and y=ED<br />2. We have EM1=EH1 ( EC is angle bisector of angle ACB)<br />DM2=DG1 (AD is angle bisector of angle BAC)<br />3. calculate FM as linear interpolation of EM1 and EM2 <br />FM=x/y*(DM2-EM1)+EM1<br />FH=x/y*(-EH1) +EH1 ( linear interpolation)<br />FG=x/y*(DG1) ( linear interpolation)<br /><br />4. FH+FG= x/y*(DG1-EH1)+EH1 <br />5. Replace DG1=DM2 and EH1=EM1 we get<br />FH+FG=x/y*(DM2-EM1)+EH1 = FM per step 3<br />Peter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-30338832165243742022010-06-29T11:32:26.596-07:002010-06-29T11:32:26.596-07:00This comment has been removed by the author.Peter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-26843967385217831242009-11-18T03:01:12.765-08:002009-11-18T03:01:12.765-08:00let A be the area and a,b,c are lenghts of the sid...let A be the area and a,b,c are lenghts of the sides BC,CA,AB. join B,F and A,F and C,F hence area of triangle ABC is equal to the sum of areas of triangles BFA,BFC,AFC. and given AD,CE are angular bisectors of angles A,C hence BE= ca/(a+b) and BD= ac/(b+c)and A =1/2ac Sin B implies Sin B = 2A/ac, and Ar of Tr BFE= 1/2 BE.FG and Ar of TrBFD= 1/2BD.FH since FG,FH are perpendiculars drawn from Fto the sides BE,BD and area of Tr BED = 1/2 (BD)(BE)(Sin B) =Aac/(a+b)(b+c). but area of Tr BED = Ar of TrBFE + Ar of Tr BFD. so by equating both L.H.S and R.H.S areas we will get the required relation that FM = FG + FHvijay9290009025http://vijay.comnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-42877334247952523312009-08-25T18:57:53.409-07:002009-08-25T18:57:53.409-07:00B:(p,q), A:(0,0) &amp; C:(b,0) AB = c, AC = b &amp...B:(p,q), A:(0,0) &amp; C:(b,0)<br />AB = c, AC = b &amp; BC = a<br />Since E divides AB in the ratio b:a<br />D:[bp/(a+b), bq/(a+b)]<br />Similarly since D divides CB in the ratio b:c<br />we&#39;ve E:[(bp+bc)/(b+c), bq/(b+c)]<br />Slope DE =((bq/(a+b))- bq/(b+c)) /(bp/(a+b)- (bp+bc)/(b+c)) = (c-a)q/(c(b-p)+a(c+p))<br />or DE is y = (c-a)qx/(cb+ac+pa-pc) + k and k is determined as = bqc/(cb+ac+pa-pc)<br />or DE is y=((c-a)qx+bqc)/(cb+ac+pa-pc)---(1)<br />Let F:(x,y) be any point on DE satisfying the eqn. above. <br />Now AB: y=qx/p or qx-py=0<br />AC: y=0<br />Slope BC = -q/(b-p)<br />BC: y= -qx/(b-p) +k1<br />Passes thru’ (b,0) so k1 = qb/(b-p)<br />Hence BC: qx+ y(b-p)-qb=0<br />Hence, FG = qx-py/c where c^2=p^2+q^2<br />FH = (qx +y(b-p)-qb)/a where a^2=(b-p)^2+q^2<br />FG and FH are of opposite signs.<br />Hence, FG + FH = (qx-py)/c - (qx +y(b-p) -qb)/a <br />If this is equated to FM = y then we obtain<br />y = ((c-a)qx+bqc)/(cb+ac+pa-pc) which is, in fact, equation (1) above. Therefore for any point F on DE we’ve FG + FH = FM .QED.<br />Ajit: ajitathle@gmail.comAjithttps://www.blogger.com/profile/00611759721780927573noreply@blogger.com