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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 1228.
AMO4 and BMO4 are congruent SSS so O4M bisects < AMB hence Tr.s AMD and BMC are congruent ASA (since AD//BC)So ABCD is a rectangle Proving that ABCD is indeed a square appears to involve tedious equations
Lets assume circle O4 has radius R. O4O3=R-R3. Let N be midpoint of AB, O4Nis perpendiclar to AB. O4O3^2 = [O4N+R3]^2 + [AE-AN]^2We get R^2-2R.R3=O4N^2+2O4N.R3+[AE-AN]^2 Also R^2=O4N^2+AN^2, We get O4N+R=(AN^2-[AE-AN]^2)/2R3 orO4N+R=AE.EB/2R3Since AE=2sqrt(R1R3),EB=2sqrt(R2R3),AB=2sqrt(R1R2)We get O4N+R=AB=CD hence ABCD is a square.
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