Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.
Click the figure below to see the complete problem 795.
Sunday, August 5, 2012
Problem 795: Intersecting Circles, Common Chord, Midpoint, Tangent, Secant Line, Perpendicular, 90 Degrees
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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
ReplyDeleteProof(I will post a link with a picture):
ReplyDeleteConnect 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°
Picture for the proof above:
ReplyDeletehttp://i1237.photobucket.com/albums/ff480/Evan_Liang/795.png
B, D, E, C are concyclic (*)
ReplyDeleteSo ∠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