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Geometry ProblemLevel: Mathematics Education, High School, Honors Geometry, College.Click the figure below to see the problem 729 details.
http://img713.imageshack.us/img713/4057/problem729.pngLet BH cut the circle at L . ( see picture)Let P is the projection of D over AC and Q is the point of symmetry of D over AC We have NH= NL ( property of orthocenter)And E, F and P are collinear ( Simson line) HLQD is isosceles trapezoid ==> N, S , P are collinear HSL is a isosceles triangle => ∠ (NHS)= ∠ (NLS)= ∠ (SQP) ( line HL // DQ)Quadrilateral DFPC is cyclic => ∠ (FPD)= ∠ (FCD) ( both angles face arc FD)But ∠ (FCD)= ∠ (BLD)= ∠ (NLS) ( angles face same arc BD)So∠ (FPD)= ∠ (NLS)= ∠ (SQP) => line EFP // line HSQ Since P is the midpoint of DQ so M will be the midpoint of HD
To follow the proof, please draw a figure:Let H’ be the orthocenter of ∆ADCLet K, J be the feet of the ⊥s BH, DH’ on ACLet KB, JE (both extended) meet at XLet DJ(extended)meet circumcircle at YJoin BY, DC. Also join XC, HYWe prove XHJD is a parallelogramwhich would imply that the diagonalsXJ, HD bisect each other.BH = DH’ (∵ each = 2 OZ, Z midpoint of AC)C, D, F, J are cyclic.∴ ∠FJD = ∠FCD = ∠BYD and so XJ ∥ BY=> XBYJ is a parallelogram (∵XB, JY each ⊥ AC)=> XB = JY = H’J (Property of orthocenter)∴ XH = XB + BH = H’J + DH’ = DJHence XHJD is a parallelogram, as required.
Refer Problem 729http://imageshack.us/photo/my-images/338/729gg.pdf/