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Geometry ProblemLevel: Mathematics Education, High School, Honors Geometry, College.Click the figure below to see the problem 759 details.
By Menelaus theorem on triangle ABC, (BD/DC)*(CF/FA)*(AE/EB)=1.Hence CF/FA = 4/1. Therefore FA:AC=1:3.Since ABC is an equilateral triangle, FA = (1/3)AC = (1/3)AB = AESo triangle AEF is an isosceles triangle. Angle EAF = 120, hence x = 30.
please disregard my previous solution. new Solution below include minor typo corrections. Apply Menelaus’s theorem to triangle ABC with secant FEDDB/DC x FC/FA x EA/EB=1 => ½ x FC/FA x ½= 1So FA/FC= ¼ => CA/CF= ¾From A draw AH //FD ( H on CB)We have CA/CF=CH/CD= ¾ => CH=3/4.CD=3/4 .2/3 .CB= ½. CBAH will be an angle bisector and altitude of triangle ABC, so x= angle(HAC)= 30
Draw EG ∥ DC (G on AC)∆s GDC and AEG are equilateral and BEGD is a parallelogram1/2 = EG/DC = FG/FC G is the midpoint of FCFG = GC = EB = DG∆FDC is rt angled at Dx = 30°
AE = BD = a/3 where AB = BC = AC = aSo BE = CD = 2a/3 Apply Menelau's Theorem to ∆ABCand transversal DEF(BE/EA).(AF/FC).(CD/DB) = 1 numericallyEasy to see BE/EA = 2 = CD/DBFollows AF/FC = 1/4, AF = a/3 = AESo x = ∠AFE = ∠AEF = (1/2)∠EAC = 30°
Let G be the midpoint of [BE].(Draw a picture yourself)From BG/BA=BD/BC =1/3 follow that triangles BDG and BAC are similar; moreover DG/CA=1/3.Now, in triangle BDE, median DG=BE/2- half of side of base, or that suffice that angle BDE is right. In fact, because triangle BDG is equilateral, follow that triangle EBD (in that order of vertices) is of type 30-60-90.The rest is simple.Visit also http://ogeometrie-cip.blogspot.com/
Let BD = AE = a . Draw DG // AB intersecting AC at G. now DC = BE = 2a. In tr.ABC, by BPT, GC = 2a and AG = a.now tr. DGC is equilateral. so DG = 2a . In tr. FGD and tr.FAE are similar. so FA = a. Now FA = AE = a. So ∠FEA = ∠AFE = x. ∠ FEA = ∠ GDF = x. In tr. DFG ∠DFG + ∠ GDF = ∠DGC. 2x = 60° or x = 30°
Easily BED is a 30-60-90 Tr. Having BE = 2.BD and the included angle 60. So < BED = 30 and hence x = 30.Sumith PeirisMoratuwa Sri Lanka