Construction of a product

Construction of an inverse

To divide an arbitrary line segment ${\displaystyle {\overline {AB}}}$ in a ${\displaystyle m:n}$ ratio, draw an arbitrary angle in A with ${\displaystyle {\overline {AB}}}$ as one leg. On the other leg construct ${\displaystyle m+n}$ equidistant points, then draw the line through the last point and B and parallel line through the mth point. This parallel line divides ${\displaystyle {\overline {AB}}}$ in the desired ratio. The graphic to the right shows the partition of a line segment ${\displaystyle {\overline {AB}}}$ in a ${\displaystyle 5:3}$ ratio.^[8]

If the midpoints of two triangle sides are connected then the resulting line segment is parallel to the third triangle side (Midpoint theorem of triangles).

If the midpoints of the two non-parallel sides of a trapezoid are connected, then the resulting line segment is parallel to the other two sides of the trapezoid.

Notation: For a triangle the vertical bars (${\displaystyle |\ldots |}$) denote its area and for a line segment its length.

Proof: Since ${\displaystyle CA\parallel BD}$, the altitudes of ${\displaystyle \triangle CDA}$ and ${\displaystyle \triangle CBA}$ are of equal length. As those triangles share the same baseline, their areas are identical. So we have ${\displaystyle |\triangle CDA|=|\triangle CBA|}$ and therefore ${\displaystyle |\triangle SCB|=|\triangle SDA|}$ as well. This yields

${\displaystyle {\frac {|\triangle SCA|}{|\triangle CDA|}}={\frac {|\triangle SCA|}{|\triangle CBA|}}}$ and ${\displaystyle {\frac {|\triangle SCA|}{|\triangle SDA|}}={\frac {|\triangle SCA|}{|\triangle SCB|}}}$

Plugging in the formula for triangle areas (${\displaystyle {\tfrac {{\text{baseline}}\cdot {\text{altitude}}}{2}}}$) transforms that into

${\displaystyle {\frac {|SC||AF|}{|CD||AF|}}={\frac {|SA||EC|}{|AB||EC|}}}$ and ${\displaystyle {\frac {|SC||AF|}{|SD||AF|}}={\frac {|SA||EC|}{|SB||EC|}}}$

Canceling the common factors results in:

(a) ${\displaystyle \,{\frac {|SC|}{|CD|}}={\frac {|SA|}{|AB|}}}$ and (b) ${\displaystyle \,{\frac {|SC|}{|SD|}}={\frac {|SA|}{|SB|}}}$

Now use (b) to replace ${\displaystyle |SA|}$ and ${\displaystyle |SC|}$ in (a): ${\displaystyle {\frac {\frac {|SA||SD|}{|SB|}}{|CD|}}={\frac {\frac {|SB||SC|}{|SD|}}{|AB|}}}$

Using (b) again this simplifies to: (c) ${\displaystyle \,{\frac {|SD|}{|CD|}}={\frac {|SB|}{|AB|}}}$ ${\displaystyle \,\square }$

Draw an additional parallel to ${\displaystyle SD}$ through A. This parallel intersects ${\displaystyle BD}$ in G. Then one has ${\displaystyle |AC|=|DG|}$ and due to claim 1 ${\displaystyle {\frac {|SA|}{|SB|}}={\frac {|DG|}{|BD|}}}$ and therefore ${\displaystyle {\frac {|SA|}{|SB|}}={\frac {|AC|}{|BD|}}}$

${\displaystyle \square }$

Assume ${\displaystyle AC}$ and ${\displaystyle BD}$ are not parallel. Then the parallel line to ${\displaystyle AC}$ through ${\displaystyle D}$ intersects ${\displaystyle SA}$ in ${\displaystyle B_{0}\neq B}$. Since ${\displaystyle |SB|:|SA|=|SD|:|SC|}$ is true, we have
${\displaystyle |SB|={\frac {|SD||SA|}{|SC|}}}$
and on the other hand from claim 1 we have
${\displaystyle |SB_{0}|={\frac {|SD||SA|}{|SC|}}}$.
So ${\displaystyle B}$ and ${\displaystyle B_{0}}$ are on the same side of ${\displaystyle S}$ and have the same distance to ${\displaystyle S}$, which means ${\displaystyle B=B_{0}}$. This is a contradiction, so the assumption could not have been true, which means ${\displaystyle AC}$ and ${\displaystyle BD}$ are indeed parallel ${\displaystyle \square }$