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| Intersection in a Quadrilateral (Posted on 2010-08-14) |
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Let E and F be the midpoints of sides BC and AD
respectively of convex quadrilateral ABCD and O the intersection of the diagonals AC and BD.
Prove that O lies inside quadrilateral ABEF if and only if
Area(AOB) < Area(COD).
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Submitted by Bractals
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Solution:
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Let ST denote the vector from point S to point T.
Let OC = -cOA and OD = -dOB, where c,d > 0.
Let P be the intersection of diagonal AC and line segment EF.
If OP = xOA, then O in ABEF <==> x < 0.
xOA = OP = yOE + (1-y)OF
=
y(OB+BE) + (1-y)(OA+AF)
=
y(OB+½BC) + (1-y)(OA+½AD)
=
y(OB+½[OC - OB]) +
(1-y)(OA+½[OD - OA])
=
½y(OB + OC) +
½(1-y)(OA + OD)
=
½y(OB - cOA) +
½(1-y)(OA - dOB)
Thus, 2x = -yc + (1 - y) and 0 = y - d(1 - y). Therefore,
x = ½(1 - cd)/(1 + d)
Therefore, O in ABEF <==> cd > 1.
Area(COD) = ½|OC×OD| =
½|(-cOA)×(-dOB)| =
cd(½|OA×OB|)
= cd Area(AOB).
Therefore, Area(AOB) < Area(COD) <==> 1 < cd <==> O in ABEF.
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