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 Medially Similar (Posted on 2005-08-19)
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.

 See The Solution Submitted by McWorter No Rating

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 Solution | Comment 1 of 3
` `
`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 2005-08-19 15:11:03

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