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.
Saturday, September 6, 2014
Geometry Problem 1041: Quadrilateral, Perpendicular, Diagonals, Congruence, Metric Relations
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It is quite obvious that ABCD is an isosceles trapezium.
ReplyDeleteThat 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.
ReplyDeleteDue to symmetric ∠(ABE)= ∠(ECD)
But ∠(MCE)= ∠(MBF)
So ∠(MCE)= ∠(ECD) => triangle DCM is isosceles and ED=EM
We have ME/ED= CG/GK => GK=3 and x=13
Good work, Jacob and Peter
ReplyDeleteIt is possible to find AD from this which is = to 21/5 and so DC = 7/10 sqrt(208)