Monday, July 21, 2008

Elearn Geometry Problem 135

Go Geometry Problem: Orthic triangle

See complete Problem 135
Orthic Triangle, Altitudes, Perpendicular, Parallel. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.


  1. 1)in triangle DEF ,ang(FEB)=90-B
    DKHG are concyclic, ang(HGK)=ang(HDK)=90-B
    ang(FEG)=ang(EGK) then EF // GK
    2)circle H is the incircle of DEF then
    triangle EFH and triangle GHK are similar
    triangle AEH and triangle DGH are similar
    GK= EF*(DG/AE)=2RsinAcosBcosC

  2. Solution to problem 135.
    1) Quadrilateral AEHF is cyclic, then
    ang(HFE) = ang(EAH) = 90 – C (1).
    Since GD and AC are parallel, then
    ang(GDH) = ang(EAH) = 90 – C.
    Quadrilateral DKHG is also cyclic, then
    ang(HKG) = ang(GDH) = 90 – C (2).
    From (1) and (2), we have ang(HKG) = ang(HFE), thus GK and EF are parallel.

    2) We have ang(BHF) = 90 – ang(ABE) = A. Quadrilateral BDHF is cyclic,
    then ang(BDF) = ang(BHF) = A.
    Furthermore, ang(BDG) = C and ang(CDK) = B, so
    ang(GDK) = A.
    Points G, N, M and K belong to the circumference with diameter HD, since these points are vertices of right angles which subtend the diameter.
    So, ang(DGK) = ang(DNK) (1).
    Furthermore, ang(DKN) = ang(BDF) = A, because they subtend the arc DGN.
    So, ang(GDK) = ang(DKN) (2).
    Triangles DGK and KND are congruent, due to (1) and (2) and because both have the common side DK.
    Thus GK = DN = DM.

  3. G,H,K,D are concyclic. They lie on a circle on HD as diameter.
    GK subtends at D, an angle = A (GD∥AC and KD∥AB).
    By Sine Rule GK = HD.sin A
    ∠NDH = ∠FDA = ∠FCA = 90° - A, So ∠NHD = A
    sin A = sin∠NHD = DN/HD, HD.sin A = DN
    So GK = DN = DM
    So GK∥FE