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

Labels:
90,
common chord,
intersecting circles,
midpoint,
perpendicular,
secant,
tangent

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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