Online Geometry theorems, problems, solutions, and related topics.
Geometry Problem. Post your solution in the comment box below.Level: Mathematics Education, High School, Honors Geometry, College.Click the figure below to view more details of problem 1240.
If < DAC = A then < DEB = A and < DOB = 2ANow DOB being isoceles < DBO = 90-ABut < DCA = 90-AIt follows that DBOC is concyclic and < CDA being 90, < BOC is also = 90Sumith PeirisMoratuwaSri Lanka
Correction to last line.......If BO meets AC at X, it follows that DBXC is concyclic and < CDA being 90, < BXC is also = 90
https://goo.gl/photos/LyXa4bZgSQcpQGuR9Draw diameter BF of circle O , lines AE and CDSince AD⊥DC= > DC will cut circle O at FSince AE⊥CE => AE will cut circle O at FIn triangle ABF , C is the orthocenter so AC⊥BO
let angle CAB = A, ABC = B and ACB = C=> angle DEC = DEB = DAC = A => triangles ABC & EBD are similarExtend BO to meet the circle at F and join FE=> <FED = 90+A Since FEDB is cyclic quadrilateral <DBF = <DBO = <ABO = 90-ASo 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