prove that the line segment joing the mid points of the adjacent sides of a quadrilatral from a parllogram.
Asked by mukul rao | 17th Jun, 2013, 10:07: PM
Let ABCD be an arbitrary convex quadrilateral (Figure 1), and let the points E, F, G and H be the midpoints of its sides AB, BC, CD and AD respectively. We need to prove that the quadrilateral EFGH is the parallelogram.
Draw the diagonals AC and BD in the quadrilateral ABCD (Figure 2). The segment HG is the midpoint segment in the triangle ACD. Therefore, the segment HG is parallel to the side AC of the triangle
ACD in accordance to the theorem: The line segment joining the midpoints of two sides of a triangle.
The segment EF is the midpoint segment in the triangle ABC. Therefore, the segment EF is parallel to the side AC of the triangle ABC. Since the segments HG and EF are both parallel to the diagonal AC, they are parallel to each other.
Similarly, the segment GF is the midpoint segment in the triangle DCB. Therefore, the segment GF is parallel to the side DB of the triangle DCB. The segment HE is the midpoint segment in the triangle ABD. Therefore, the segment HE is parallel to the side DB of the triangle ABD. Since the segments GF and HE are both parallel to the diagonal DB, they are parallel to each other.
Thus, we have proved that in the quadrilateral EFGH the opposite sides HG and EF, HE and GF are parallel by pairs. Hence, the quadrilateral EFGH is the parallelogram.
Answered by | 18th Jun, 2013, 02:20: AM
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