Wednesday, February 15, 2012

Problem 729: Triangle, Altitude, Orthocenter, Circumcircle, Perpendicular, Midpoint

Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the problem 729 details.

Online Geometry Problem 729: Triangle, Altitude, Orthocenter, Circumcircle, Perpendicular, Midpoint.



    Let 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

  2. To follow the proof, please draw a figure:
    Let H’ be the orthocenter of ∆ADC
    Let K, J be the feet of the ⊥s BH, DH’ on AC
    Let KB, JE (both extended) meet at X
    Let DJ(extended)meet circumcircle at Y
    Join BY, DC. Also join XC, HY
    We prove XHJD is a parallelogram
    which would imply that the diagonals
    XJ, 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’ = DJ
    Hence XHJD is a parallelogram, as required.

  3. Refer Problem 729