Dividing Evenly a Quadrilateral II
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Copyright © 1996-2012 Alexander BogomolnyIt appears that the problem of bisecting the area of a quadrilateral ABCD by a line through a vertex (C below) has been proposed by Professor W. McWorter as a ruler and compass construction in 1964. In 2004, it has been used at the Rasor-Bareis-Gordon competition at the Ohio State University. The applet reflects Prof. McWorter's solution.
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A line parallel to AC is drawn through either B (to the intersection with AD) or through D (to the intersection with AB.) CM is the median of either ΔCDE or ΔCBE. CM bisects the area of the relevant triangle and, where M falls inside a side of ABCD (AD or AB as the case may be), it also divides the area of quadrilateral ABCD. For the construction to go through, it needs to be shown that one of the two cases always takes place. One way of doing that is by following the previous construction: any line (in particular that parallel to AC) through the midpoint K of the base BD of ΔABD crosses one of the sides, either AB or AD. This point serves as the midpoint of the side opposite vertex C in one of the triangles CDE or CBE.
Area of Quadrilateral
- Brahmagupta's Formula and Theorem
- Carpets in a Quadrilateral
- Carpets in a Quadrilateral II
- Dividing Evenly a Quadrilateral
- Dividing Evenly a Quadrilateral II
- Area of a Bicentric Quadrilateral
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Copyright © 1996-2012 Alexander Bogomolny| 40620154 |
