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Place a triangle in the coordinate plane. Use the midpoint formula to find the coordinates of the midpoints and then find the slope of the relevant segments.
See solution.
Let's consider a triangle ABC with vertices A(0,0), B(a,0), and C(b,c). Also, let S and T be the midpoints of AC and BC, respectively.
Our mission is to prove that ST is parallel to AB. Before drawing ST, let's find their coordinates by using the Midpoint Formula.
| M(x_1+x_2/2,y_1+y_2/2) | |
|---|---|
| A(0,0) and C(b,c) | B(a,0) and C(b,c) |
| S(0+b/2,0+c/2) or S(b/2,c/2) | T(a+b/2,0+c/2) or T(a+b/2,c/2) |
Next, let's connect S and T.
Substitute ( (0,0)) & ( (2a,0))
Substitute ( (b,c)) & ( (a+b,c))
Given: & △ ABC & S is the midpoint ofAC & T is the midpoint ofBC Prove: & ST∥ AB Proof: Let A(0,0), B(a,0), and C(b,c) be the vertices of △ ABC. The Midpoint Formula allows us to find the coordinates of S and T, which are ( b2, c2) and ( a+b2, c2), respectively. Next, we calculate the slope of ST which is given by c/2-c/2(a+b)/2-b/2 or 0. Lastly, the slope of AB is 0-0a-0=0.
Because the slope of ST and AB are equal, we conclude that ST∥ AB.