Let P,Q and R be the three vertices of a triangle. Prove that the line
segment that joints the midpoints of two of the sides is parallel to and half the length
of the third side.
Solution
in your triangle ABC, R is the mid point of AB and S is the mid point of AC. Now the basic
proportionality theorem recalls the fact that when a line segment is drawn parallel to the base of
the triangle say RS drawn parallel to BC, then AR/RB = AS/AC Using the converse of basic
proportionality theorem, we get AR/AB = 1 as R is the mid point of AB. And AS/AC is also
equal to 1 as S is the mid point of AC. Hence AR/RB = AS/AC establishes by converse of BPT
theorem, RS becomes parallel to AB. So first part of the problem has been proved. ie the line
segment drawn is || to the third side. For second part, as AR/AB = 1/1 by ratio and
proportionality we get AR /(AR+RB = 1/2 ====> AR /AB = 1/2 Same way, we can prove
AS/AC = 1/2 NOw in the two triangles ABC and ARS, by SAS rule the two triangles become
similar. Hence RS / AB = 1/2 Or RS = AB/2 So second part is also proved. ie the segment is half
of the third side.