Geometry Problem. Post your solution in the comments box below.

Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the complete problem 1031.

## Sunday, July 20, 2014

### Geometry Problem 1031: Trapezoid, Intersecting Circles, Common Chord, Tangent Line, Midpoint, Parallel Lines

Subscribe to:
Post Comments (Atom)

See sketch for location of S, L, N

ReplyDeleteWe have SE^2=SA.SB and SF^2=SC.SD

SB/SA=SC/SD

1. Calculate p^2= (SE.SD)^2= SA.SB.SD^2

And p1^2=(SA.SF)^2= SC.SD.SA^2

(p/p1)^2= (SB/SA).(SD/SC)= 1

So p=p1 => quadrilateral AFED is cyclic

Similarly quadrilateral BFEC is cyclic

2. We have ∠(FAE)= ∠(FDE)= ∠(BFC)= ∠ BEC) => AE//FC and FD//BE

And FNEL is a parallelogram

We also have NE.NA=NF.ND => N will be on radical line GH of circles O and Q

Similarly L will be on GH

M is the intersection of 2 diagonals of a parallelogram FNEL so M is the midpoint of EF.

To Peter Problem 1031

DeleteSketch link?

below is the sketch of problem 1031

ReplyDeletehttp://s25.postimg.org/xlwtf6fjz/pro_1031.png

Peter Tran

link is broken

Delete