### The SaRD Construction:

An Elegant Solution for Euclid's Partitioning Problem

By

**SriRanga S. Dattatreya**

(with help from Ravi E. Dattatreya)

December 14, 1996

Summit, New Jersey

*Requirement*.

Partition a line segment into a given number of equal parts.

*Construction.*

Let *AB* be the given line segment. Draw a line
segment *AC* such the *AC* is perpendicular to *AB*
at the point *A*. Draw a line *L* as shown, such
that L is perpendicular to *AB* at *B*. (Note: AC
and the line L do not have to be perpendicular to AB, as long as AB and
L are parallel.)

Now, mark a point *D* on *L* such that the length
of the segment *BD* is *(n-1)* times the length
of the segment *AC*. Also, ensure that *D* and *C*
lie on opposite sides of the line segment *AB*. Draw the segment
*CD*. The segments *AB* and *CD* intersect
at *P.*

AP is the required partition. Length *AP *= *(1/n)*
length *AB.*

### Geometric Proof:

See Figure 1. Triangles *CAP* and *DBP*
are similar, because:
Angle* A* = angle *B* = 90 by construction

Angle *APC* = angle*BPD*, vertically opposite angles

The ratio *CA:BD* is equal to the ratio *AP:PB*,
by similarity of the triangles.

Ratio *CA:BD* is equal to *1:(n-1)* (by construction).

Thus, *AP:PB* is equal to *1:(n-1)* (from result
above).

Or, *AP:AB* is *1:n. *(middle school algebra)

*Q.E.D.*

### Algebraic *Proof:*

See figure 2. Let the point *A* be the origin, *(0,0)*.
Let the length *AB* be equal to *b*, so that point
*B* is at *(b,0).* Let the lenght *AC*
be *a*, so that the point *C* is at *(0,a)*
and the point *D* is at *(b, -(n-1)a)*, by construction.
The line *CD* meets the *x*-axis at *P*.
*X*-intercept is *AP*.* *The line *CD*
meets the *y*-axis at *C*. *Y*-intercept
is *AC*.

The slope of the line *CD* is *[a - (-(n-1)a)] / [0 - b],*
using basic principles. This simplifies to *-(na/b).* The *y*-intercept
of this line is *a*, at *C*. Thus, the equation
of this line is *y = (na/b) x + a,* in the standard *y
*= *mx *+ *c* format. To get the *x*-intercept,
set *y* to zero to get,* x = -a / (na/b) = - ab/na = -b/n.
*That is, the distance *AP* is *b/n*. Since
the distance *AB* is *b*, the point *P*
represents the required *nth* partion.

*Q.E.D.*

Why do we prefer **SaRD** Construction to the **GLaD** Construction?

**SaRD** is much more simple to construct geometrically.

**SaRD** gives the student *insight* as to why it works.

**SaRD** is not an iterative procedure, unlike **GLaD**.

**SaRD** reminds one of Euclid's own construction.

**SaRD** can handle non-integral partitions, e.g., *1/(2.5),*
as long as we can mark off the appropriate length BD to be *1.5*
times the length *AC.*

SaRD can also handle multiples. For example, to find a point *E*
on the line *AB* (extended) such that *AE* is *n*-times
the length *AB*, then, choose the point *D* to be
on the same side as *C*, and make *BD* equal to
*(1-(1/n)) *of *AC*.

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