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Divide in Half (Posted on 2014-03-07) Difficulty: 3 of 5

Given a convex quadrilateral. Construct, with straight-edge and
compass, a line parallel to one of the sides of the quadrilateral
that divides the area of the quadrilateral in half.

See The Solution Submitted by Bractals    
Rating: 5.0000 (1 votes)

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Some Thoughts re(3): Response to Solution | Comment 5 of 6 |
(In reply to re(2): Response to Solution by Bractals)

Here’s a possible way of constructing the bisector if it turns out to cross
two adjacent sides. (Using square brackets to denote polygonal areas).
Label the quadrilateral so that BA and CD meet when produced, and so
that AD and BC meet at O when produced.
Use the method already posted to construct a line PQ, parallel to AB, with
P on AO and Q on BO, so that [ABQP] = [ABCD]/2.
If PQ crosses two adjacent sides of the quadrilateral then the following
further steps are needed.
Let X be the intersection of CD and PQ, and let the required bisecting line
be YZ with Y on CD and Z on BC.
To compensate for the ‘lost’ triangle PXD, we need [XQZY] = [PXD].

Considering triangles with the same vertical angles:

                        [PXD]/[ CXQ] = (|PX||DX|)/(|CX||QX|)

Therefore:         [CYZ]/[CXQ] = 1 - (|PX||DX|)/(|CX||QX|)

i.e.                    (|CY|/|CX|)2  = 1 - (|PX||DX|)/(|CX||QX|)

Thus                             |CY|2 = |CX|*(|CX|  -  |DX||PX|/|QX|)

|DX||PX|/|QX| can be constructed by dividing DX in the ratio ||PX|:|QX|.

|CY| can be found as the tangent length to a circle. Hence Y and YZ.

So YZ is constructible and a similar, but modified method can be used for
bisectors parallel to all four sides.

There must be a simpler way – but I can’t think of it.

  Posted by Harry on 2014-03-25 21:32:12

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