Online Geometry theorems, problems, solutions, and related topics.
Geometry ProblemLevel: Mathematics Education, High School, Honors Geometry, College.Click the figure below to see the complete problem 672.
Extend AD to meet circle O again at ECircles O and O' touch each other at ASo points A, O', O are collinearOA being a diameter(of circle O'),ODA is a right angleSo OD bisects the chord ADE in circle ONow a.b = AD.DB = AD.DE = AD^2Hence AD = sqrt(a.b)
Extend AD to meet circle O at E. Since OA is a diameter of circle O',OD is perpendicular to AD or OD is perpendicular chord AE of circle O. In other words D is the midpt. of AE which means AD*DC = AD*DE = AD^2 (intersecting chords) or a*b = AD^2 or AD is the GM of a & b.Ajit : firstname.lastname@example.org
Typo Correction:BD * DC = AD * DE = AD^2 since AD = DEAjit
Let AD extended meet circle O at P< ODA = 90 so PD = DA = xHence easily in circle O,PD. DA = a.b i.e. x^2 = a.bSumith PeirisMoratuwaSri Lanka