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

Labels:
diagonal,
parallelogram,
perpendicular,
similarity,
triangle

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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