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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 1041.
It is quite obvious that ABCD is an isosceles trapezium. That is, it is a cyclic quadrilateral. On the other hand, AC⊥BD and EF⊥AB. By Brahmagupta theorem, EF bisects CD. Since FG//HK, thus G is also the mid-point of CK. Hence, GK=3, and x=BK=7+3+3=13.
Draw CM //EG. M is on BE.Due to symmetric ∠(ABE)= ∠(ECD)But ∠(MCE)= ∠(MBF)So ∠(MCE)= ∠(ECD) => triangle DCM is isosceles and ED=EMWe have ME/ED= CG/GK => GK=3 and x=13
Good work, Jacob and PeterIt is possible to find AD from this which is = to 21/5 and so DC = 7/10 sqrt(208)