## 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.

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

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°

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

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