Online Geometry theorems, problems, solutions, and related topics.
Geometry Problem. Post your solution in the comments box below.Level: Mathematics Education, High School, Honors Geometry, College.Click the figure below to see the complete problem 1029.
Extend DE to F so that ED=EF => BDCF is a parallelogramNote that ∠(FDC)= ∠(DFB)= ∠(DBA) => BA is a tangent to circumcirle of triangle DBFCircumcenter of DBF will be on BC => E is the circumcenter of DBFSo EB=ED=EC => BD perpendicular to AC
Is this solution is correct? :1)Continue DE till point F such that BE = CE = EF --> BFC right triangle and E is the circumscribed circle of BFC and since BC is the diameter And angle ABC is right so AB is tanget and so on ....
Hi AntonioPlease publish full of my problema.Prob 1027 We build AH perpendicular BC and DF perpendicular AB CH=HD=1/2DC=1/2ED DF=DH=1/2ED α =10 X=90-4α X=50 Prob 1028 BE=ED=EC EM=EN EN =1/2 EC Triangle ECN < ECN =30 ECN = 90-4X -> X =15 Prob 1029 We note <BDF = 90 BD perpendicular AC Erina New Jersey
as stated, problem 1029 is wrong.Given right ABC and BE=CE: ∠(DBA)=∠(EDC) does not imply BD⊥ACTake D' on AC such as ED'⊥BC, then ∠(D'BA)=∠(ED'C) but BD' is not ⊥ACbleaug
To Bleaug Problem 1029 There is a condition CD greater than AD,Thanks
naahhh, seriously?then you should add this in the "Given:" sectionbleaug
To Bleaug Now it has been included in the "Given" section. Thanks
Thanks Antonio for the clarification.Above solutions do not make explicit use of this newly added premiss, therefore they must be incomplete at the very least. Without it, there are two solutions to the problem, only one obeying CD>AD. For example in Peter's solution, there is no justification of the implication on the 3rd line.Background: if we make D vary on AC and let F on ED be such that ∠(EDC) = x = ∠(FBA), it is easy to show that ∠(BFE) is constant and equal to ∠(BAC). See http://bleaug.free.fr/gogeometry/1029.pngTherefore, the locus of all F when D varies lies on the circle circumscribed to FEB. This circle crosses AC in at most two points. With simple geometric arguments we can show that the two following points belong to both AC and circle FEB and thus can be candidate solutions: D1 such as BD1⊥AC and D2 midpoint of AC. Since AD>CD, only D1 is acceptable => BD⊥ACbleaug
Draw FE perpendicular to BC to meet BD extended at F. Since AB // FE < EDC = < ABF = < BFE = < CFE since FE is the perpendicular bisector of BC. So CEDF is cyclic and BD is thus perpendicular to ACSumith PeirisMoratuwaSri Lanka
Another method - let GE be the perpendicular bisector G on AC. Show that < CAB = ABG = < BGE = < CGE. Hence < DEG = < DBG , hence BDGE is cyclic and the result followsSumith PeirisMoratuwaSri Lanka