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Geometry Problem. Post your solution in the comments box below.Level: Mathematics Education, High School, Honors Geometry, College.Click the figure below to view the complete problem 1046.
EA²+EC² = AC² = 4R²EB²+ED² = BD² = 4R²EA²+EB²+EC²+ED² = 8R²
By the symmetry of the figure, B, O & D are collinear making BD & AC diameters of the circumcircle each = 2R. And EB²+ED²=BD²=4R². Likewise, EA²+EC²=4R². Hence the result.
the solution is obtained by applying the Pythagorean theorem to the diagonals AC and BD AExAE + ECxEC = ACxAC = 4 (rxr) BExBE + EDxED = BDxBD = 4 (rxr) then AExAE +ECxEC + BExBE +EDxED = 8 (rxr)
EA^2+ EC^2 =AC^2=4R^2;EB^2+ ED^2 =BD^2=4R^2=>EA^2+ EC^2+EB^2+ ED^2=2.4R^2=8R^2
we have two times the application of the Pythagorean theorem applied to the two diagonals of the rectangle Square and add them give exactly 8 (rxr)
The result follows easily from applying Pythagoras to right Tr.s AEC and BED