Prove the following theorem is true: “If a segment joins the midpoints of two sides of a triangle, then the s

Question

Prove the following theorem is true: “If a segment joins the midpoints of two sides of a triangle, then the segment is parallel to the third side and is half its length.”

http://img.photobucket.com/albums/v225/warpedtrojan/05_08h_05a.gif

Given
R is the midpoint of OP
S is the midpoint of QP

Prove
RS is parallel to OQ
RS=.5(OQ)


Thanks!

Answer 1

Extend the triangle into the paralellogram OPUQ, where OP and OQ are sides of the paralellogram and PQ is the shorter of its two diagonals. Then the triangles RPS and STQ are congruent, therefore RS=ST=1/2 of RT which equals OQ. Also, by construction RS is clearly parallel to OQ.

Answer 2

Because of the Talete's theorem,
if we take 3 parallel lines (P, RS',OQ ), with R being the middle point of PO then PS' and S'Q will be "similar" to PR and RO. that said PS' would be equal to S'Q, so we know that the line parallel to OQ passing by R divides PQ in two equal parts.
This is the segment we were working on!

Answer 3

the new segment divides the original triangle into a triangle with a vertex that is common to the original triangle, and two sides of half the length double the size of this new triangle, and it will have the same SAS as the original, and all the ASA relationships will also be preserved this tells us that the ratios of all sides is preserved, so the third side of the smaller triangle is 1/2 the size of the original triangle's 3rd side since the ASA relationships are preserved, the 3rd side of both triangles must be parallel

Answer 4

If R is the midpoint of segment OP, then R = ((0+b)/2, (0+c)/2)
= ( b/2, c/2).
If S is the midpoint of segment QP, then R = ((a+b)/2, (0+c)/2)
= ((a+b)/2, c/2).

The slope of segment RS is (c/2 - c/2)/((a+b)/2 - b/2) = 0.
The slope of segment OQ is (0 - 0)/(a - 0) = 0.

If slopes are equal, then segment RS is parallel to segment OQ.

The length of segment RS is sqrt((a+b)/2 - b/2)^2 + (c/c - c/2)^2)
= sqrt((a+b-b)/2)^2 + 0^2) = sqrt((a/2)^2) = a/2.

The length of segment OQ is sqrt((a - 0)^2 + (0 - 0)^2) = sqrt(a^2) = a.

RS = a/2
RS = (OQ)/2
RS = 0.5(OQ)

Answer 5

Extend RS to the left to the point T such that RS = ST. Draw QT. It shoud now be easy for you to show that ORTQ is a parallelogram and hence OR||OQ. It should also be obvious that RS = 1/2 OQ