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

Labels:
90,
circle,
circumcircle,
diameter,
midpoint,
perpendicular,
tangent,
triangle

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See diagram

ReplyDeletehere.Since 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.