(In reply to Response to Solution
The plot thickens.
I had assumed that we were to construct the dividing line parallel to any given side, even if
it crossed two adjacent sides. So I’ve been stuck on what I thought was the harder part of
Having missed the boat on this one, I’ll attach what I’d done anyway (very similar to the posted
solution) and have a think about your additional challenge. Much fun.
Let XY be the required line, being parallel to side AB of a quadrilateral ABCD, with X and Y
on lines AD and BC respectively.
If AD and BC are parallel, draw a line parallel to AB through the mid-point of CD to form a
parallelogram. XY can then be drawn through the centre of this parallelogram.
Otherwise, let O be the point of intersection of AD and BC, and use the convention that
any point, P say, will be at distance p from O. Thus |OA| = a etc.
Considering the area of triangles with a common angle at O, the area bisection by XY leads to
xy = (1/2)(ab + cd) and, since XY is parallel to AB, x/a = y/b which can be combined to give:
x2 = (a/2)(a + cd/b) (1)
Draw CE parallel to BD to intersect OA at E. Triangles OEC ~ ODB => e = cd/b
Now construct a point F on OA produced such that AF = e. (This can be done using a
parallelogram OE’A’A where E’ is any point with |OE’| = |OE|). Thus f = a + cd/b.
Let G be the mid-point of OA so that g = a/2.
Equation (1) shows that x is the length of a tangent from O to a circle with diameter FG.
Draw this circle (with centre H say) and draw a further circle with diameter OH,
denoting one of their intersections by T. Since /OTH is a right angle, OT is a tangent to
the first circle and its length, x, will be the radius of a circle centre at O that crosses
OA at X. The line XY can then be drawn parallel to AB.
* If X and Y lie on the line segments AD and BC then all is well. If one of them lies
outside its respective segment then we have further work to do, since the required line cuts
two adjacent sides rather than opposite sides...
Posted by Harry
on 2014-03-18 12:14:54