tag:blogger.com,1999:blog-6933544261975483399.post8939757637279022230..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Geometry Problem 1127: Triangle, Circumcircle, Tangent, Circle, Diameter, Midpoint, Perpendicular, 90 DegreesAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger1125tag:blogger.com,1999:blog-6933544261975483399.post-71098394945949429772021-08-02T09:22:42.767-07:002021-08-02T09:22:42.767-07:00See diagram here.
Since F is on BD, AF ⊥ FB =>...See diagram <b><a href="https://drive.google.com/file/d/1_6_boK5Nuq7MBkQNVeb8zjgpIQshtsW6/view?usp=sharing" rel="nofollow"> here</a></b>.<br /><br />Since F is on BD, AF ⊥ FB => AF ⊥ FD => ΔAFD is right in F.<br /><br />∠AFD = ∠AMD = π/2 => A, M, F, D are concyclic (O3, circumcircle of ΔAFD).<br /><br />On circle O : AD tangent to O in A => ∠HAD = ∠HCA = ∠HCE (1)<br /><br />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).<br /><br />Hence, A, M, H, F, D are all on the same circle O3.<br /><br />Now, ∠CHM = 2.π - ∠CHF - ∠FHM.<br /><br />Using circles O2 and O3 : ∠CHM = 2.π - ∠CEF – (π - ∠FAM).<br /><br />Hence ∠CHM = ∠FAM + (π - ∠CEF) = ∠FAE + ∠FEA = π/2 QED<br /><br />Discussion :<br />This proof is for the configuration of the problem as per the given diagram.<br />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).<br />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.<br />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.<br />Greghttps://www.blogger.com/profile/14941282981782772132noreply@blogger.com