Proof of Right Triangle Midpoint Vertex Ratio Theorem

The ratio of the size of a line drawn from the midpoint of the hypotenuse of a right triangle to the vertex opposite the hypotenuse is 1:2.

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Proof of Right Triangle Midpoint Vertex Ratio Theorem To show:
  1. Draw a right triangle. Label the vertex of the right angle 'C', another vertex 'A', and the remaining vertex 'B'. Note that the non right angle vertexes can be selected in any order. This makes the proof general for any right triangle.
  2. Draw a midpoint on the hypotenuse. Lable the midpoint 'D'.
  3. Draw a midpoint on side AC. Label the midpoint 'E'.
  4. Draw a line from E to D. Note that ED is parallel to CB. How do we know they are parallel?
  5. Draw a midpoint on side BC. Label the midpoint F.
  6. Draw a line from F to D. Note that FD is parallel to AC. How do we know they are parallel?
  7. Since ED is parallel to CB and CB is perpendicular to AC, we can conclude that ED is perpendicular to AC. A similar argument can be made to show DF is perpendicular to CB.
  8. The measure of EC is 1/2 of AC (property of a segment with endpoints on midpoints), and the measure of CF is 1/2 CB (by the same argument). Since EDFC is a rectangle (How do we know that EDFC is a rectangle?), we can conclude that ED = CF, so ED is 1/2 of CB.
  9. By the Pytahgorean Theorem, EC^2 + ED^2 = CD squared. Also, AC^2 + CB^2 = AB^2.
  10. By substitution, (2EC)^2 + (2ED)^2 = AB^2 --> 4(EC^2) + 4(ED^2) = AB^2 --> EC^2 + ED^2 = 1/4(AD^2) --> EC^2 + ED^2 = (1/2AD)^2 = CD^2 --> 1/2 AD = CD, which was to be shown. Why do we use only the positive root?