The theorem of de Ceva once more

The theorem of de Ceva once more

(See also EWD989, where the theorem was erroneously attributed to Gupta.)

We identify the row vector $p = \left( {{p}{0},{p}{1},{p}{2}}\right)$ with the line $p$ whose equation is ${p0} \times + {p1} \cdot y + {p2} = 0$ ,and similarly for $q$ and $r$ . We identify the column vector $P = \left( {{R}{x},{P}{y},1}\right)$ with the point $P$ ,whose $x,y$ -coordinates are $\left( {{R}{x},{P}{y}}\right)$ ,and similarly for $Q$ and $R$ . When the equations are normalized,a scalar product like $q \cdot R$ equals ${R}{s}$ perpendicular distance from $q$ .

We now consider lines $p, q, r$ and points $P, Q, R$ , such that $P$ lies on $p$ , $Q$ lies on $q$ and $r$ goes through $R$ , i.e.

(0) ${p}^{ \cdot }P = 0\;{9}^{ \cdot }Q = 0\;r \cdot R = 0$

and now observe

(1) $\left( {p \cdot Q}\right) \cdot \left( {q \cdot R}\right) \cdot \left( {r \cdot P}\right) + \left( {p \cdot R}\right) \cdot \left( {q \cdot P}\right) \cdot \left( {r \cdot Q}\right) = 0$

$$ \begin{array}{l} d e t \left| \begin{array}{c c c} 0 & p \cdot Q & p \cdot R \ q \cdot P & 0 & q \cdot R \ r \cdot P & r \cdot Q & 0 \end{array} \right| = 0 \ \equiv {0 } \ \end{array} $$

$$ d e t \left| \begin{array}{l l l} p \cdot P & p \cdot Q & p \cdot R \ q \cdot P & q \cdot Q & q \cdot R \ r \cdot P & r \cdot Q & r \cdot R \end{array} \right| = 0 $$

[\left{ \begin{array}{ll} \text{ definition of product of matrices}\end{array}\right} ]

$$ d e f \left(\left| \begin{array}{l} p \ q \ r \end{array} \right| \times \left| \begin{array}{l} P Q R \end{array} \right|\right) = 0 $$

$\Rightarrow$ {determinant of product is

$$ \det \left| \begin{array}{l} p \ q \ r \end{array} \right| = 0 \quad \vee \quad \det \left| P Q R \right| = 0 $$

[\left{ \begin{array}{ll} \text{ geometrical } & \text{ interpretation } \ & \text{ other } \end{array}\right} ]

lines $p,q,r$ are concurrent or parallel)

(poinrs $P,Q,R$ are collinear)

The usual formulation of the theorem of de Ceva states in the context of a nondegenerate triangle PQR (in our terminology)

lines $p,q,r$ are concurrent

$$ (\frac {p \cdot Q}{p \cdot R}) \cdot (\frac {q \cdot R}{q \cdot P}) \cdot (\frac {r \cdot P}{r \cdot Q}) = - 1 $$

but

• this formula is only an implication and not an equivalence

both formulation and proof break down if some factors in (1) equal zero - the quotient $p \cdot Q/p \cdot R$ is formulated as $\overline{AQ}/\overline{AR}$ ,where $A$ is the point of intersection of $p$ and ${QR}$ ;the case that $p$ and ${QR}$ are parallel is thus ruled out, and would require a separate proof. - the case that lines $p,q,r$ are parallel is traditionally ignored (also when the converse theorem is derived that concludes concurrency from (1)).

In short: three cheers for calculation!

Nuenen, 3 June 1997

prof. dr. Edsger W. Dyrkstra Department of Computer Sciences The University of Texas at Austin Austin, TX 78712 - 1188 USA