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

Labels:
congruence,
diagonal,
metric relations,
perpendicular,
quadrilateral,
triangle

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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)