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

Labels:
altitude,
orthic triangle,
parallel,
perpendicular,
tangency point,
triangle

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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