Geometry Problem. Post your solution in the comments box below.
Level: Mathematics Education, High School, Honors Geometry, College.
Click the diagram below to enlarge it.
Saturday, July 4, 2015
Geometry Problem 1127: Triangle, Circumcircle, Tangent, Circle, Diameter, Midpoint, Perpendicular, 90 Degrees
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See diagram here.
ReplyDeleteSince F is on BD, AF ⊥ FB => AF ⊥ FD => ΔAFD is right in F.
∠AFD = ∠AMD = π/2 => A, M, F, D are concyclic (O3, circumcircle of ΔAFD).
On circle O : AD tangent to O in A => ∠HAD = ∠HCA = ∠HCE (1)
Using line DFE, circle O2 and (1) : ∠HFD = π - ∠HFE = π - ∠HCE = π - ∠HAD => A, H, F, D are concyclic (also O3, circumcircle of ΔAFD).
Hence, A, M, H, F, D are all on the same circle O3.
Now, ∠CHM = 2.π - ∠CHF - ∠FHM.
Using circles O2 and O3 : ∠CHM = 2.π - ∠CEF – (π - ∠FAM).
Hence ∠CHM = ∠FAM + (π - ∠CEF) = ∠FAE + ∠FEA = π/2 QED
Discussion :
This proof is for the configuration of the problem as per the given diagram.
However, it can easily be shown that the problem is undefined when ΔAFD is right in B (in which case the tangents to A and C are parallel and D doesn’t exist) or isoscele in B (in which case E, F and M are one and the same point so that O2 is not defined).
These two degenerate configurations define 4 non degenerate sets of configurations, depending on whether B is acute or obtuse and angle A is greater or smaller than angle C.
The corresponding 3 other proofs are derived from this one by adapting the equation governing the angles ∠HFD and ∠HAD to prove the concyclicity of A, F, M, H, D, on the one hand, and that governing the angles ∠CHM, ∠CHF and ∠FHM, on the other hand, but are otherwise identical to that for the given diagram.