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 1009.

## Friday, April 25, 2014

### Geometry Problem 1009: Tangent Circles, Common External Tangent, Common Internal Tangent, Arithmetic Mean

Labels:
arithmetic mean,
circle,
common tangent

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Since EA=ET=EB, FC=FT=FD,

ReplyDeleteAlso AD//BC (ABCD is trapezium),

Thus EF=(AD+BC)/2.

Join the points CT and DT.

ReplyDeleteAlso join the points AT and BT.

We have

DF=FT=FC=DC/2

From the similarity between triangles DFT and DTA we get

DT/AD=DC/2DT

(1.)2DT^2=(AD)(DC)

And from the similarity between triangles CFT and CTB we get

CT/BC=DC/2CT

(2.)2CT^2=(BC)(DC)

Adding equation 1 and 2 we get

2(DT^2+CT^2)=(AD+BC)DC

Because DTC is a right triangle we have

2DC^2=(AD+BC)DC

2DC=(AD+BC)

I forgot the add the final piece:

DeleteDC=EF because AB=DC and therefore DC=2DF=2AE then

2EF=AD+BC

EF=(AD+BC)/2