Pappus’s theorem


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.


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,


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



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.


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}|


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


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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