Math Forum - Project of the Month, February 1997

# February POM - Winner

## Justin Lam Grade 8 Sequoia Middle School, Pleasant Hill, California

From: quan.lam@ucop

Let the quadrilateral be ABCD. Let E, F, G, H be the midpoints of sides AB, BC, CD, DA respectively. AC and BD are the two diagonals.

Since E and F are midpoints of Triangle ABC, EF//AC. Similarly, HG//AC because H and G are midpoints of Triangle DAC. So, EF//HG.

Using the same argument for Triangles ABD and CBD, we have EH//BD and FG//BD. So, EH//FG.

Therefore, for any quadrilateral, the quadrilateral formed by connecting the consecutive midpoints must be at least a PARALLELOGRAM with the length half the length of the corresponding diagonals.

1. If the two diagonals AC and BD are of equal length, then the parallelogram would be equilateral or a RHOMBUS.
2. If the two diagonals are perpendicular to each other, then the parallelogram would be a RECTANGLE.
3. Of course, if the diagonals are both equal length and perpendicular to each other, the parallelogram would be a SQUARE.

• Rectangle - the parallelogram is equilateral or a Rhombus (case 1)
• Kite - it is a rectangle (case 2)
• Isosceles Trapezoid - Rhombus (case 1)
• Square - Square (case 3)
• Rhombus - Rectangle (case 2)