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
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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