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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 1221.
Since EF//CD => Arc(EC)=Arc(FD) In circle O , We have ∠ (DTB)= ½( Arc(EC)+Arc(DB))= ½(Arc(FD)+Arc(DB))= ½ Arc(FB)= ∠ (BAF)… (1)Since DT tangent to circle Q at T => ∠ (BAT)= ∠ (DTB)…. (2)Compare ( 1) to (2) we have ∠ (BAF)= ∠ (BAT)So F, T , A are collinear
< DTB = < FET but < DTB = < TABSo < FET = < TAB But < FET = < FABHence < TAB = < FABSo, F, T, A are collinearSumith PeirisMoratuwaSri Lanka
Is arc EC=arc FD. Suppose that AT intersects the circle with center O to the point Κ,then <EKA=<EBA=<ATC=<KTD so EK//CD.Therefore arc KD=arc FD, so the points K, F will coincide.Therefore points F,T,A are collinear.
Consider the line FA intersects the circle Q at P. Join AB. P and T coincide and hence F,T,A are collinear