See complete Problem 135
Orthic Triangle, Altitudes, Perpendicular, Parallel. Level: High School, SAT Prep, College geometry
Post your solutions or ideas in the comments.
Monday, July 21, 2008
Elearn Geometry Problem 135
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1)in triangle DEF ,ang(FEB)=90-B
ReplyDeleteDKHG are concyclic, ang(HGK)=ang(HDK)=90-B
ang(FEG)=ang(EGK) then EF // GK
2)circle H is the incircle of DEF then
DN=DM=(DE-EF+FD)/2=(R/2)*(-sin2A+sin2B+sin2C)
=2RsinAcosBcosC
triangle EFH and triangle GHK are similar
triangle AEH and triangle DGH are similar
GK= EF*(DG/AE)=2RsinAcosBcosC
Solution to problem 135.
ReplyDelete1) 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.
G,H,K,D are concyclic. They lie on a circle on HD as diameter.
ReplyDeleteGK 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
∠EGK=∠HGK=∠HDK=∠HCD=∠HED=∠BED=∠BAD=∠FAH=∠FEH=∠FEG,
So GK∥FE