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Geometry ProblemLevel: Mathematics Education, High School, Honors Geometry, College.Click the figure below to see the complete problem 781.
A trigonometric Proof: (sketch)Let∠DMB=θIn ΔOBM:R/cosθ = R cosB/sin(90°–θ–C+A) = R cosB/cos(C–A+θ),cosB cos θ = cos(C-A)cosθ – sin(C–A)sinθ,(*) tanθ = 2cosC. cosA/sin(C–A)In ΔBDM: DM/cosA = R cosB/ sin(θ+90°–A) = Rcos B/cos(A – θ)In ΔBEM: ME/cosC = RcosB /sin(θ–(90°–C))= - Rcos B /cos(C+θ)It suffices to prove: cosA cos(C+θ)+cosC.cos(A-θ)= 0 (or) cosθ(2cos C.cos A)= sinθ.sin(C – A) This follows from (*)Hence DM = ME
Let N be the midpoint of AB. Then NM || AH and ON || CH so angle(ONM)=angle(C1HA)=B.But ONDM is inscrided, then angle(ONM)=angle(ODM)=B.The same is true for angle OEM. So DO=DE.
So DO=EO, therefore OM is the perpendicular bisector of DE and M is the midpoint of DE.
AnonymousYou got an excellent solution. !attached is the sketch per your solutionhttp://img9.imageshack.us/img9/3192/problem781.pngPeter Tran