Sunday, August 5, 2012

Problem 795: Intersecting Circles, Common Chord, Midpoint, Tangent, Secant Line, Perpendicular, 90 Degrees

Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the complete problem 795.

Online Geometry Problem 795: Intersecting Circles, Common Chord, Midpoint, Tangent, Secant Line, Perpendicular, 90 Degrees.

4 comments:

  1. Let KL is perpendicular to AB. Then, FBGE, CBDE and KBHF are all subscribed-able quadrilaterals. So, BH is perpendicular to CD and from this MH is perpendicular to FG. You can see this here: http://mtz256.files.wordpress.com/2012/08/7953.jpg

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  2. Proof(I will post a link with a picture):
    Connect CB,BD.
    By a property of tangent lines:
    ∠ECD=∠CBA,∠EDC=∠ABD
    ∵180-∠CED=∠ECD+∠EDC
    ∴180-∠CED=∠CBA+∠ABD=∠CBD
    ∴C,B,D,E are concyclic points
    ∵∠EFB=∠EGB=90°
    ∴E,F,B,G are also concyclic points
    Connect BE
    By properties of concyclic figures:
    ∠BED=∠DCB, ∠BEG=∠BFG
    ∴∠DCB=∠BFG
    ∴C,F,H,B are concyclic points
    Connect BH
    ∴∠CHB=∠CFB=90°
    To prove that MH⊥FG, we just need to prove that ∠FHC=∠MHB
    By properties of concyclic figures:
    ∠FHC=∠FBC,∠HBF=∠FCH=∠CBA
    ∴∠HBA=∠CBF=∠FHC
    ∵△AHB is a right triangle and M is that midpoint of AB.
    ∴∠MHB=∠HBA=∠FHC (Done)
    ∴∠FHC+∠CHM=∠MHB+∠CHM=90°

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  3. Picture for the proof above:
    http://i1237.photobucket.com/albums/ff480/Evan_Liang/795.png

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  4. B, D, E, C are concyclic (*)
    So ∠BDC =∠BEC
    B, G, E, F are concyclic
    So ∠BEF = ∠BGF
    Follows
    ∠BDH = ∠BDC = ∠BEC = ∠BEF = ∠BGF= ∠BGH
    So B, G, D, H are concyclic; also BGD = 90⁰
    Hence ∠BHD = 90⁰
    (*)∠CBD =∠ CBA + ∠DBA = ∠ECD +∠ EDC = 180⁰ - ∠ CED

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