Online Geometry theorems, problems, solutions, and related topics.
Geometry Problem. Post your solution in the comment box below.Level: Mathematics Education, High School, Honors Geometry, College.Click the figure below to view more details of problem 1237.
https://goo.gl/photos/YYPhTktUP43uz8Pt5Denote (XYZ) as angle (XYZ)Let θ= (FBA)=(FAB)=(FAD)=(DAE)=(DCF)=(EDA) ( see sketch)So (BFC)=2 θ ( external angle)(CMX)=2 θ … ( MX//AE)(MXD)=2 θ(CDX)= θLet H is the projection of E over AFLet u=BF=FA , v= DA=DC , w=ED=EA and t=MC=MFIn triangle BCF we have FC= u/cos(2 θ)= 2.tAD=v= AC/2.cos(θ)= u/(2.cos(θ)) x ( 1+1/cos(2. θ))….. (1)AE=w= v/(2.cos(θ))…. (2)Replace (1) into (2) and simplifying we get AE=w= u/(2.cos(2θ))= t=AF/(2.cos(2θ)) => AF= 2.AE.cos(2. θ)= 2.AHSo H is the midpoint of AF and triangle AEF is isosceles (EAF)= (EFA)= 2. θ => B, F, E are collinear we also have EF=EA=FM=MX => triangle MFE is isosceles => ME is an angle bisector of ( BEX)In parallelogram MXEA , since DE=FM= w=t => XD=FA=FB So XE=BE => triangle XEB is isosceles X and D are symmetric points of B and F with respect to symmetric axis EM So XF will intersect BD at a point on the symmetric axis EM
Problem 1237Let's say that <FBA=<FAB=<FAD=<DCA=<EAD=<EDA=xThe straight CD intersects the extension of BF in point K.Then <CFB=2x=<FCK+<FKCso <FKC=x and CF=FK.Then triangle FAK=triangle FBC.So <FAK=90. Passes the midpoint E’ of FK,is triangleFBM=triangleFAE’ so <E’AF=2x or E’AD=x.But triangles FBA and DCA are similarthen BA/AC=FA/DA so triangles CBA and DFA are similar.Then <FDA=<BCA=y.But <FDC=180-<DCF-<DFC=180-(x+<FAD+<FDA)=180-(2x+y)=180-90=90.So MD=MF=MB=MC=FE’=AE’=KE’=DE’.Is triangleADE=triangleADE’,so the points E, E’ coincides.Ιs <EDA=<CAD=x so XE//MA therefore MXEA is parallelogram with <ΜΧD=<MAE=2x=<MDX.Then triangleMXD=triangleMBF soXD=BF and triangleFDX=triangleFDB.Si BD intersect FX in point P then the point P It lies on the perpendicular bisector of FD.Byt ME is perpendicular bisector of FD.Therefore lines BD,FX and ME are concurreant.APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE