Geometry Problem. Post your solution in the comments box below.
Level: Mathematics Education, High School, Honors Geometry, College.
This entry contributed by Ajit Athle.
Click the figure below to enlarge the problem 937.
Sunday, December 1, 2013
Geometry Problem 937: Triangle, Circumcircle, Circles, Tangent, Sides, Midpoint
Subscribe to:
Post Comments (Atom)
http://img41.imageshack.us/img41/4263/7fbb.png
ReplyDeleteDraw points M, L and F per attached sketch.
We have right angles BCL, BCF, BAM, BAF, CBM and ABL.
So M, A and F are collinear . L, C and F are collinear
And BM// FL and BL//MF => MBLF is a parallelogram and diagonals BF and ML cut at midpoint O of BF
Since OaOc is a perpendicular bisector of BD => M, D and L are collinear and OD perpen. To BE
So D is the midpoint of BE
OOc is perpendicular to AB and OOa is perpendicular to BC so <OcOOa=180-<B. OcBOa=OcDOa=180-<B as well, so OcOaDO is cyclic. BOaOOc is parallelogram so OOc and DOa are both radii of circle Oa. That makes OcOaDO an isosceles trapezoid with DO parallel to OcOa both perpendicular to BD. Therefore D must be midpoint of chord BE.
ReplyDeleteApply inversion on B in a way such that line AC is interchanged with circum circle O
ReplyDeleteLine BA and line BC are unchanged whilst circle Oa and circle Oc becomes straight line through A and C , parallel to BC and BA respectively.
So the inversion image forms parallelogram B-A-C-D, as D is the intersection of the two inverted lines.
Note that point E is inverted along BDE and now it is lying on AC, which is the intersection of diagonal of the parallelogram.
The Remaining of proof is trivial by inversion property .
QED