Three drawing lines almost meet. Is that drawing error or an exact property? This additional lesson reviews four distinct questions: concurrence, collinearity, lengths on a circle and triangle centers. It is not a prerequisite for the main course.
Prerequisites: medians and altitudes, cyclic quadrilaterals, similarity, area.
Ceva: three cevians meet
In nondegenerate triangle ABC let D,E,F lie strictly inside sides BC,CA,AB. Lines AD,BE,CF meet at one point if and only if:
(BD)/(DC) ⋅ (CE)/(EA) ⋅ (AF)/(FB) = 1.
These are positive length ratios. BD/DC=2 and CE/EA=2 require AF/FB=1/4, not 1/2. With A(0,0), B(9,0), C(0,6), choose D(3,4), E(0,2), F(9/5,0). All three lines pass through P(9/7,12/7).
Why a product? If P is the intersection, BD/DC equals the area ratio PAB/PCA, CE/EA equals PBC/PAB, and AF/FB equals PCA/PBC, by equal heights in the relevant triangles. Their product cancels to 1. Conversely, intersect the first two cevians: the required ratio specifies a unique internal F on AB, forcing the third line through P.
Ceva: internal points and product +1
Menelaus: side extension, product −1
Ptolemy: cyclic 8×6 rectangle
Euler: O, G, H on y=2
Menelaus: three points lie on a line
Now D,E,F may lie anywhere on side lines BC,CA,AB, excluding vertices. Define the directed segment [UV] as coordinate V−U on its line, oriented B→C, C→A, A→B. Reversing endpoints reverses the sign. Internal division gives a positive ratio, external division a negative one.
Menelaus’s criterion: D,E,F are collinear if and only if ([BD]/[DC])·([CE]/[EA])·([AF]/[FB])=−1. Keep D(3,4), E(0,2) in the previous triangle but choose F(−3,0). Line DE is y=2x/3+2 and contains F. The product is 2·2·(−1/4)=−1.
Unsigned lengths would give +1 here; the sign records a side extension. Do not mix conventions. Ceva above intentionally uses internal points: in the general directed version, product +1 also allows three parallel cevians, without a finite common intersection.
Ptolemy: diagonals and opposite sides
For a convex cyclic quadrilateral with distinct vertices A,B,C,D named in circular order:
AC ⋅ BD = AB ⋅ CD + BC ⋅ DA.
Our 8×6 rectangle has diagonals 10: 10·10=8·8+6·6, consistent with Pythagoras. A rectangle is cyclic; an arbitrary quadrilateral need not be. Keep the circle condition and vertex order. The equality checks lengths, not concurrence.
Euler’s line: three centers
In a nonequilateral triangle the circumcenter O, centroid G and orthocenter H are collinear, with OG:GH=1:2. For an equilateral triangle all three coincide and determine no unique line. The incenter generally does not belong to this list.
For A(0,0), B(8,0), C(2,6), we get O(4,2), G(10/3,2), H(2,2). All lie on y=2; OG=2/3, GH=4/3. Coordinates verify our example; the general theorem needs a proof.
Branch · With formulasOptional · prerequisite: vectors G021
A short vector proof
Take O as origin. Vertex vectors a,b,c have equal length R. For h=a+b+c, (h−a)·(b−c)=(b+c)·(b−c)=|b|²−|c|²=0. This point lies on the altitude from A, and likewise the others, so it is H. The centroid vector g=(a+b+c)/3 gives h=3g: G lies between O and H, dividing OH in ratio 1:2. Exclude the equilateral case for nonzero distances. Here h denotes a point vector, not height.
Identify what needs proving, then check the theorem’s conditions. An almost-correct drawing is no substitute for an equality or proof of concurrence.
Sources and limits
Merzliak and Yakir, Geometry grade8, additional sections, printed pp.116–119,126–127,199–200; exact PDF pages are recorded in research. Independent treatments: Wayne Bishop, Modern Geometry §4.3 for Ceva/Menelaus; University of St Andrews for Ptolemy; AMSI for Euler’s line. Examples, coordinate drawings and checks are our own; the vector branch uses G021.