Question

prove that the straight line joining the mid points of the sides of an isosceles trapezium in order form a rhombus

Answer 1

I have labeled the trapezoid ABCD with AB and CD parallel, and |AD|=|BC|. The midpoints of the sides are F,G,H and I as in the diagram 1.

I want to show that FGHI is a rhombus, that is that the four sides are of equal length. My proof is in three steps.

Step 1:

Angle ADC = Angle DCB

Proof: (kindly see figure 2)

Draw a line through B parallel to AD meeting DC at E. ADEB is a parallelogram so |AD|=|BE| which is also |BC|. Thus triangle BEC is isosceles and angle BEC = angle DCB. Since AD is parallel to BE angle ADC = angle BEC. Thus Angle ADC = Angle DCB.

Step 2:

|AC|=|DB|

Proof: (see figure 3, 4)

Using step 1 and side-angle-side, triangles ADC and BDC are congruent.

Proof of the main result:

Join A to C. In triangle ABC the line FG joins the midpoints of two of the legs so |AC|=2|FG|. Similarly, using triangle ADC, |AC|=2|IH|.

If you now join B to D, a similar argument shows that |BD|=2|IF|=2|HG|. Since, by step 2, |AC|=|DB|, |FG|=|IH|=|IF|=|HG|.