Geometry Problem. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.
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Saturday, July 30, 2016
Geometry Problem 1240: Triangle, Circle, Diameter, Circumcircle, Circumcenter, Perpendicular. Mobile Apps
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If < DAC = A then < DEB = A and < DOB = 2A
ReplyDeleteNow DOB being isoceles < DBO = 90-A
But < DCA = 90-A
It follows that DBOC is concyclic and < CDA being 90, < BOC is also = 90
Sumith Peiris
Moratuwa
Sri Lanka
Correction to last line.......
DeleteIf BO meets AC at X, it follows that DBXC is concyclic and < CDA being 90, < BXC is also = 90
https://goo.gl/photos/LyXa4bZgSQcpQGuR9
ReplyDeleteDraw diameter BF of circle O , lines AE and CD
Since AD⊥DC= > DC will cut circle O at F
Since AE⊥CE => AE will cut circle O at F
In triangle ABF , C is the orthocenter so AC⊥BO
let angle CAB = A, ABC = B and ACB = C
ReplyDelete=> angle DEC = DEB = DAC = A => triangles ABC & EBD are similar
Extend BO to meet the circle at F and join FE
=> <FED = 90+A
Since FEDB is cyclic quadrilateral <DBF = <DBO = <ABO = 90-A
So if a triangle is formed by extending AC to intersect BO at a point P - it is right angle at P as <PAB = A & PBA = 90-A
Let (Q) be the circle with diameter AC, and Bt the tangent to circle (O) on B.
ReplyDeleteCircles (Q) and (O) intersect at D and E.
Line (ADB) cuts (Q) on A and (O) on B.
Line (ECB) cuts (Q) on C and (O) on B. Applying Reim’s theorem:
AC//Bt and since Bt┴BO BO┴AC.