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Let-ABCD-be-a-convex-quadrilateral-and-let-E-and-F-be-the-points-of-intersections-of-the-lines-AB-CD-and-AD-BC-respectively-Prove-that-the-midpoints-of-the-segments-AC-BD-and-EF-are-collinear-

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Question Number 16068 by Tinkutara last updated on 21/Jun/17
Let ABCD be a convex quadrilateral and let E and F be the points of intersections of the lines AB, CD and AD, BC, respectively. Prove that the midpoints of the segments AC, BD, and EF are collinear.
LetABCDbeaconvexquadrilateralandletEandFbethepointsofintersectionsofthelinesAB,CDandAD,BC,respectively.ProvethatthemidpointsofthesegmentsAC,BD,andEFarecollinear.
Answered by ajfour last updated on 30/Jun/17
z_I =(z_1 /2) ; z_G =((mz_2 +z_1 +k(z_2 −z_1 ))/2) ; z_G −z_I =(−(k/2))z_1 +(((m+k)/2))z_2 z_D =z_1 +λ(z_1 −mz_2 ) also z_D =μ[z_1 +k(z_2 −z_1 )] z_D −z_D =0 ⇒ (1+λ−μ+μk)z_1 −(λm+μk)z_2 =0 ⇒ 1+λ−μ(1−k)=0 and μ=−((λm)/k) so 1+λ+((λm(1−k))/k)=0 k+λ(k+m−mk)=0 or λ=(k/(mk−(m+k))) z_D = z_1 +(k/([mk−(m+k)]))(z_1 −mz_2 ) =((mkz_1 −mz_1 −kz_1 +kz_1 −mkz_2 )/(mk−(m+k))) =((m(k−1)z_1 −mkz_2 )/(mk−(m+k))) z_H =((z_2 +z_D )/2) ⇒ z_H −z_I =((z_2 +z_D −z_1 )/2) z_H −z_I =((z_2 −z_1 )/2)−((m(k−1)z_1 −mkz_2 )/(2(mk−m−k))) =(((mk−m−k)(z_1 −z_2 )−mk(z_1 −z_2 )+mz_1 )/(2(mk−m−k))) =((−kz_1 +(m+k)z_2 )/(2c)) ∀ c=mk−m−k z_H −z_I =(1/c)[−(k/2)z_1 +(((m+k)/2))z_2 ] =((z_G −z_I )/c) ⇒ z_H ,z_G , and z_I are collinear.
zI=z12;zG=mz2+z1+k(z2−z1)2;zG−zI=(−k2)z1+(m+k2)z2zD=z1+λ(z1−mz2)alsozD=μ[z1+k(z2−z1)]zD−zD=0⇒(1+λ−μ+μk)z1−(λm+μk)z2=0⇒1+λ−μ(1−k)=0andμ=−λmkso1+λ+λm(1−k)k=0k+λ(k+m−mk)=0orλ=kmk−(m+k)zD=z1+k[mk−(m+k)](z1−mz2)=mkz1−mz1−kz1+kz1−mkz2mk−(m+k)=m(k−1)z1−mkz2mk−(m+k)zH=z2+zD2⇒zH−zI=z2+zD−z12zH−zI=z2−z12−m(k−1)z1−mkz22(mk−m−k)=(mk−m−k)(z1−z2)−mk(z1−z2)+mz12(mk−m−k)=−kz1+(m+k)z22c∀c=mk−m−kzH−zI=1c[−k2z1+(m+k2)z2]=zG−zIc⇒zH,zG,andzIarecollinear.
Commented by ajfour last updated on 30/Jun/17
let me see your book′s solution..
letmeseeyourbook′ssolution..
Commented by ajfour last updated on 30/Jun/17
Answered by Tinkutara last updated on 30/Jun/17
Let P, Q, and R be the midpoints of AC, BD, and EF. (Figure). Denote by S the area of ABCD. As we have seen the locus of the points M in the interior of ABCD for which [MAB] + [MCD] = (1/2) S is a segment. We see that P and Q belong to this segment. Indeed, [PAB] + [PCD] = (1/2)[ABC] + (1/2)[ACD] = (1/2) S. and [QAD] + [QCD] = (1/2)[ABD] + (1/2)[BCD] = (1/2) S. Now we have [RAB] = (1/2)[FAB], since the distance from F to AB is twice the distance from R to AB. Similarly, [RCD] = (1/2)[FCD]. We obtain [RAB] − [RCD] = (1/2)[FAB] − [FCD] = (1/2) S. Taking into account the observation in the solution to Problem 16064, it follows that P, Q and R are collinear.
LetP,Q,andRbethemidpointsofAC,BD,andEF.(Figure).DenotebyStheareaofABCD.AswehaveseenthelocusofthepointsMintheinteriorofABCDforwhich[MAB]+[MCD]=12Sisasegment.WeseethatPandQbelongtothissegment.Indeed,[PAB]+[PCD]=12[ABC]+12[ACD]=12S.and[QAD]+[QCD]=12[ABD]+12[BCD]=12S.Nowwehave[RAB]=12[FAB],sincethedistancefromFtoABistwicethedistancefromRtoAB.Similarly,[RCD]=12[FCD].Weobtain[RAB]−[RCD]=12[FAB]−[FCD]=12S.TakingintoaccounttheobservationinthesolutiontoProblem16064,itfollowsthatP,QandRarecollinear.
Commented by Tinkutara last updated on 30/Jun/17

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