Question 999912
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Let two triangles &nbsp;{{{DELTA}}}{{{A[1]}}}{{{B[1]}}}{{{C[1]}}}&nbsp; and &nbsp;{{{DELTA}}}{{{A[2]}}}{{{B[2]}}}{{{C[2]}}} &nbsp;are similar, &nbsp;so that the pairs of their corresponding sides &nbsp;{{{A[1]}}}{{{B[1]}}}&nbsp; and &nbsp;{{{A[2]}}}{{{B[2]}}}, 

&nbsp;{{{B[1]}}}{{{C[1]}}}&nbsp; and &nbsp;{{{B[2]}}}{{{C[2]}}}, &nbsp;and &nbsp;{{{A[1]}}}{{{C[1]}}}&nbsp; and &nbsp;{{{A[2]}}}{{{C[2]}}} &nbsp;are proportional with the same &nbsp;(common) &nbsp;coefficient of proportionality.

Now consider two corresponding medians &nbsp;{{{A[1]}}}{{{D[1]}}}&nbsp; and &nbsp;{{{A[2]}}}{{{D[2]}}}. &nbsp;Consider the triangles &nbsp;{{{DELTA}}}{{{A[1]}}}{{{D[1]}}}{{{C[1]}}}&nbsp; and &nbsp;{{{DELTA}}}{{{A[2]}}}{{{D[2]}}}{{{C[2]}}}. 

They have two pairs of proportional sides &nbsp;{{{A[1]}}}{{{C[1]}}}&nbsp; and &nbsp;{{{A[2]}}}{{{C[2]}}}, &nbsp;{{{D[1]}}}{{{C[1]}}}&nbsp; and &nbsp;{{{D[2]}}}{{{C[2]}}}&nbsp; with the same coefficient proportionality. 

For the last pair, &nbsp;{{{D[1]}}}{{{C[1]}}}&nbsp; and &nbsp;{{{D[2]}}}{{{C[2]}}}, &nbsp;it is true because these segments are halves of the corresponding sides &nbsp;{{{B[1]}}}{{{C[1]}}}&nbsp; and &nbsp;{{{B[2]}}}{{{C[2]}}}.

The angles <I>L</I>{{{C[1]}}} and <I>L</I>{{{C[2]}}}&nbsp; between these proportional sides are congruent, &nbsp;as they are corresponding angles of the similar original triangles. 

Thus the triangles &nbsp;{{{DELTA}}}{{{A[1]}}}{{{D[1]}}}{{{C[1]}}}&nbsp; and &nbsp;{{{DELTA}}}{{{A[2]}}}{{{D[2]}}}{{{C[2]}}}&nbsp; have two pairs of proportional sides and the congruent angles between them. 

According to the &nbsp;<U>SAS-test of similarity for triangles</U>, &nbsp;these triangles are similar.

Therefore, &nbsp;their sides &nbsp;{{{A[1]}}}{{{D[1]}}}&nbsp; and &nbsp;{{{A[2]}}}{{{D[2]}}}&nbsp; are proportional with the same coefficient of proportionality. 

It is exactly what has to be proved, &nbsp;since &nbsp;{{{A[1]}}}{{{D[1]}}}&nbsp; and &nbsp;{{{A[2]}}}{{{D[2]}}}&nbsp; are the corresponding medians of the original triangles. 
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