## Pappus’s theorem

Problem:

Given a point on the circumference of a cyclic quadrilateral, prove that the product of the distances from the point to any pair of opposite sides or to the diagonals are equal.

Proof:

Let a, b, c, d be the coordinates of the vertices A, B, C, D of the quadrilateral and consider the complex plane with origin at the circumcenter of ABCD. Without loss of generality, assume that the circumradius equals 1.

The equation of line AB is

$\left | \begin{array}{ccc} a & \overline{a} & 1 \\ b & \overline{b} & 1 \\ z & \overline{z} & 1 \end{array} \right | = 0$.

This is equivalent to $z(\overline{a}-\overline{b})-\overline{z}(a-b)=\overline{a}b-a\overline{b}$, that is,

$z+ab\overline{z}=a+b$

Let point $M_{1}$ be the foot of the perpendicular from a point M on the circumcircle to the line AB. If m is the coordinate of the point M, then

$z_{M_{1}}=\frac{m-ab\overline{m}+a+b}{2}$

and

$d(M, AB)=|m-m_{1}|=|m-\frac{m-ab\overline{m}+a+b}{2}|=|\frac{(m-a)(m-b)}{2m}|$ since $m \overline{m} = 1$.

Likewise,

$d(M, BC)=|\frac{(m-b)(m-c)}{2m}|$, $d(M, CD)=|\frac{(m-c)(m-d)}{2m}|$

$d(M, DA)=|\frac{(m-d)(m-a)}{2m}|$, $d(M, AC)=|\frac{(m-a)(m-c)}{2m}|$

and $d(M, BD)=|\frac{(m-b)(m-d)}{2m}|$

Thus,

$d(M, AB).d(M, CD)=d(M, BC).d(M, DA)=d(M, AC).d(M, BD)$ as claimed.

QED.

More later,

Nalin Pithwa

PS: The above example indicates how easy it is prove many fascinating theorems of pure plane geometry using the tools and techniques of complex numbers.

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