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Geometry ProblemLevel: Mathematics Education, High School, Honors Geometry, College.Click the figure below to see the complete problem 701.
∠ACB = 90° and CD ⊥ ABAlso OD ⊥ EF and bisects it (DE = DF)∴ CD² = AD.DB = DE.DF = DE² = DF²So ∠ECF = 90° = ∠ACBSubtracting common ∠FCB from both sides,We get ∠ACF = ∠BCE
Note that m(ACB)=90 and triangle EOF is isosceles . OD is an altitude of tri. EOF ,so DE=DFIn Circle O, power from point D is DA.DB=DE.DF=DE^2 (1)In circle O’, power from point D is DA.DB=DC^2 (2)From (1) and (2) we get DC=DE=DF and m( ECF)=90Both angles( ACF) and ( BCE) supplement to angle (BCF) so m(ACF)=m(BCE)Peter Tran