perplexus.info :: Geometry : Quadrilateral Equals Overlapping Triangles

Quadrilateral Equals Overlapping Triangles (Posted on 2010-07-23)


Let P and Q be points on the sides AB and CD respectively
of convex quadrilateral ABCD. Prove the following:

    |AP|     |CQ|
If ------ = ------, then [ABCD] = [QAB] + [PCD]
    |PB|     |QD|

where [KLM...XYZ] denotes the area of polygon KLM...XYZ.

See The Solution Submitted by Bractals

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Solution

Comment 1 of 1

Let |AP|/|PB| = |CQ|/|QD| = u/v where u + v = 1.

The perpendicular distance of Q from AB can be expressed as a weighted mean of the perpendicular distances of D and C from AB, by using u and v as the respective weights. Thus [QAB] can also be written as a weighted mean as follows:
                                                [QAB] = u[DAB] + v[CAB]
and, by similar reasoning:           [PCD] = u[BCD] + v[ACD]
Adding these two equations gives

            [QAB] + [PCD]   = u([DAB] + [BCD]) + v([CAB] + [ACD])
                                    = u[ABCD] + v[ABCD]
                                    = (u + v)[ABCD]
                                    = [ABCD]                      as required.

Posted by Harry on 2010-07-26 20:59:54

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