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Given a convex polygon with 4 or more sides, is there a way to transform (convert, reduce) that polygon to a triangle having the same area as the polygon by using only a compass and straight edge?

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You may be interested in the Bolyai-Gerwien Theorem. All the constructions can be done with compass and straightedge. – André Nicolas Sep 29 '12 at 16:10

Yes, and it is not necessary that the polygon is convex.

Divide the polygon into triangles. For each triangle, construct a rectangle with the same base and half the height, so that the triangle and the rectangle have the same area.

For each rectangle with sides $a$ and $b$, construct a square of side $c$, where $c$ is a mean proportional of $a$ and $b$. (This means that $a:c=c:b$.) This is done by drawing a semicircle with diameter $a+b$, and and constructing a perpendicular to the diameter from the point where the segments of length $a$ and $b$ meet. Then $c$ is the height of the perpendicular inside the semicircle.

Given two squares with sides $c$ and $d$, construct a right triangle with sides $c$ and $d$, then the hypothenuse will have length $\sqrt{c^2+d^2}$. Hence we can construct a single square with area $c^2+d^2$.

By iterating this construction, we end up with a square of the same size as the original polygon.

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For a tetragon $ABCD$, draw one of its diagonal, say $BD$, and put the orthogonal lines to this from the other two vertices, then add these heights $m_A$ and $m_C$ on one of these sides, along the orthogonal line, say the line of $m_A$, yielding a point $A'$. Then triangle $BCA'$ has the same area as the tetragon.

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