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.

Posted by Bractals
on 20050819 15:11:03 