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Geometry ProblemLevel: Mathematics Education, High School, Honors Geometry, College.
Click the figure below to see the complete problem 660.
Here’s a proof by brute force. Draw a line // DF from A meeting EF & GD in M & N resply. Now by similar triangles, AM=GE*AF/FG while AN =AG*DF/FG. Now let A be the origin, AC the x-axis and AM, the y-axis so that we can write eqns. for EF & GD as follows:x/AF+y*FG/(GE*AF)= 1 and x/AG+y*FG/(AG*DF)=1 from which we can eliminate y to obtain the x-co-ordinate of H the intersection of EF & GD as x= (GE*AF–AG*DF)/(GE-DF.Now let /_DAC= α=90–C and /_CAE=β=180–A. Hence we can say that:x=[b^2 sinβcosβ(cosα)^2-b^2 sinαcosα(cosβ)^2]/(sinβcosβ-sinαcosα)=2bcosβcosαsin(β-α)/(sin(2β)–sin(2α))=bsinC*cosA/sin(B) using standard trigonometric formulae.Further, c/sinC = b/sinB by the sine rule hence, x = c*cosA. In other words, the x-co-ordinate of H is the same as that of B or HB is perpendicular to CA, our x-axis.Ajit
http://img220.imageshack.us/img220/3329/problem660.pngHere is the geometry solution.Connect EC and AD.EG cutt DC at M and DF cut AE at NLet H’ and B’ are the projections of H and B over ACNote that (ADF)=(ACD) and (GCE)=(FNA) (see picture)Triangles AFD similar to triangle GEC and AF/FD=GM/GC ( 1)Triangles AFN similar to triangle EGC and AF/FN=GE/GC ( 2)Divide expressions (1) to (2) we have FN/FD=GM/GE (3)From (3) we also get FD/GE=ND/ME=kTriangle HFD similar to triangle HEM with ratio of similarity = k so H’F= k/(1-k) x FGTriangle BND similar to triangle BEM with ratio of similarity = k so B’F= k/(1-k) x FGH’ coincide with B’ and HB perpendicular to ACPeter Tran
Extend DA, CE to meet at X. Join BX.Extend CA to meet BX at Y.AD ⊥ BC and CE ⊥ ABFollows X is the orthocentre of ∆ABC.So BYX ⊥ ACLet Z be the midpoint of AC.Clearly D, Y, E and Z are concyclic (∵they lie on the ninepoint circle of ∆ABC)So ∠DYZ = ∠DEZ = ∠EDZ (∵DZ = EZ)= ∠EYZSame as ∠DYF = ∠EYGFollows ∆DYF ∼ ∆EYGHence YF/YG = DF/GE = HD/HG (∵FD ∥ GE)Follows HY ∥ DF or HY ⊥ ACAlready BY ⊥ ACHence HB ⊥ AC
Hallo Peter Tran, can you exlain why From (3) we also get FD/GE=ND/ME=k. Thank's
To AnonymousSee details below. From (3) FN/FD=GM/GE we have FN/GM=FD/GE=(FN+FD)/(GM+GE)=ND/ME (Properties of ratio)Peter Tran