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Geometry ProblemLevel: Mathematics Education, High School, Honors Geometry, College.Click the figure below to see the complete problem 910.
∠KDJ= ∠KDA + ∠ADJ= ∠KDA + ∠ABJ= 1/2 [∠CDA + ∠ABC]= 90°⇒ KJ is diameter of the circle O⇒ OK = OJ = RIn ΔOIK, IK² = R² + d² - 2Rd cos∠IOKIn ΔOIJ, IJ² = R² + d² - 2Rd cos∠IOJBut ∠IOK + ∠IOJ = 180°, cos∠IOK = −cos∠IOJAdding up, we have IK² + IJ² = 2 (R² + d²)
KI^2=R^2+d^2-2*R*d*cos<KOI, and IJ^2=R^2+d^2+2*R*d*cos<KOI. Adding the two equations gives the result.