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Geometry ProblemLevel: Mathematics Education, High School, Honors Geometry, College.Click on the figure below to enlarge it.
Enlarge circle Q with center E, such that Q→O, (Homothetic transformation)let D→F, then EQO, COF and EDF are straight lines, with EF=7. Since then CF is the diameter of circle O, thus ∠CED=∠CEF=90°. Using Pythagoras theorem, we have CE² = 4²−3² = 7CF² = CE² + EF² = 7 + 49 = 56So OE = OF = √14Since ΔEQD~ΔEOF, x/√14 = 3/7x = (3/7)√14
If <BDE=a then <OED=90-a and <EOD=2a-90. By circle theorem if <EOD=2a-90 then <BCE=a-45 and <ECO=(a-45)+45=a=<EDB making CEDO cyclic. That makes <DEC=90 and CE=sqrt7 by Pythagoras. If QD cuts CE at P then Q is cicrumcenter of PED so QD=1/2*DP. By angle bisector PE=3/sqrt7 so DP=6*sqrt(2/7) by Pythagoras. QD is half so QD=3*sqrt(2/7)