Let ABC be a triangle, with M the midpoint of BC, and DEF a triangle, with N the midpoint of EF. Suppose that angle BAM equals angle EDN and angle CAM equals angle FDN. Show that triangles ABC and DEF are similar.
Clearly, <BAC = <BAM + <CAM = <EDN + <FDN = <EDF.
Using the sine rule,
|BA| sin(<BAM) = |BM| sin(<BMA)
|AC| sin(<CAM) = |CM| sin(<AMC)
Since |BM| = |CM| and <BMA + <AMC = 180,
|BA| sin(<BAM) = |AC| sin(<CAM)
Using the sine rule,
|ED| sin(<EDN) = |EN| sin(<END)
|DF| sin(<FDN) = |FN| sin(<DNF)
Since |EN| = |FN| and <END + <DNF = 180,
|ED| sin(<EDN) = |DF| sin(<FDN)
or
|ED| sin(<BAM) = |DF| sin(<CAM)
Therefore,
|BA| |ED|
------ = ------
|AC| |DF|
Hence, triangles ABC and DEF are similar.
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Posted by Bractals
on 2005-08-19 15:11:03 |
Solution
