Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.
Click the figure below to see the complete problem 836.
Tuesday, December 25, 2012
Problem 836: Parallelogram, Perpendicular, Diagonal, Similarity, Metric Relations
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Using Pythagoras Theorem,
ReplyDeleteEB=3, EC=√65, AD=BC=√65-3
Let G be a point such that EAGC is a rectangle.
Then CD=5, DG=3.
Since AFGC concyclic,
AD×DG=FD×DC
FD=3/5×(√65-3)
Using Pythagoras Theorem again, in ΔADF,
AF=4/5×(√65-3)
Now since AECF concyclic, by Ptolemy Theorem,
AC×EF = AE×CF + CE×AF
9x = 4/5×[3√65+16] + 4/5×[65-3√65]
9x = 4/5×81
x = 36/5
SOLUTION by Michael Tsourakakis
ReplyDeleteBecause AB //CD, is ,FA perpendicular of AB.
Let is ,CH perpendicular of AB. Then , AFCH is rectangle. So,the EHCA is, cyclic quadrilateral, additional, AECF is, cyclic quadrilateral, therefore , angle AHF=angle ACF=angleAEF,so, EHFA is, cyclic quadrilateral, , therefore , angle EHA=angleECA=angleCAD
Then, the triangles AEF and ACD, are similar.So, x:9=4:5 therefore x=36:5
see the image http://img26.imageshack.us/img26/8026/p836parallelogramperpen.gif
Let AC, BD intersect at O.
ReplyDeleteThe circle on AC as diameter passes through E, F.
Its centre is O and diameter 2R = 9
The angle θ subtended by its chord EF at C = ∠ECD = ∠EBA
x = Length of the chord EF = 2Rsinθ (by sine rule)
= 9(4/5) = 36/5