Cevians & Mass Points

Why the two products look alike. Both theorems measure how three points divide the three sides, going around the triangle $B \to C \to A \to B$. For Ceva, each ratio $\tfrac{BD}{DC}$ equals the area ratio $\tfrac{[ABP]}{[ACP]}$ of the two triangles on either side of the cevian, and the three area ratios multiply to $1$. For Menelaus, drop perpendiculars from $A$, $B$, $C$ to the transversal: each ratio is a ratio of two of those three distances, and again everything cancels. The formulas agree; the pictures (one point versus one line) do not, which is why you decide by the picture.

Why it works. Masses $m_B$ and $m_C$ balance at the point $D$ with $m_B \cdot BD = m_C \cdot DC$, so the heavier mass sits nearer the fulcrum. Once two feet are balance points, the centre of mass of all three masses lies on both cevians, hence at their intersection $P$; and on cevian $AD$ it balances $m_A$ at $A$ against $m_B + m_C$ at $D$. That lever is the ratio $AP : PD = (m_B + m_C) : m_A$.

Why it works. Area is half base times height. Two triangles with the same apex and bases on one line have the same height (the distance from the apex to that line), so their areas are in the ratio of their bases; two triangles on the same base are in the ratio of their heights. Every area argument in this topic is one of these two facts, applied once per step.

Spotting it. Three cevians through one point: Ceva gives the third foot. The ratio ALONG $CF$ is a mass-point question. Doing both is also a check, since the masses must reproduce Ceva's answer.

Step 1 (Ceva). $\dfrac{BD}{DC} \cdot \dfrac{CE}{EA} \cdot \dfrac{AF}{FB} = \dfrac13 \cdot \dfrac12 \cdot \dfrac{AF}{FB} = 1$, so $AF : FB = 6 : 1$.

Step 2 (masses). $BD : DC = 1 : 3$ gives $m_B = 3m_C$; $CE : EA = 1 : 2$ gives $m_C = 2m_A$. Take $m_A = 1$, $m_C = 2$, $m_B = 6$. Then $F$ balances $A$ and $B$ at $AF : FB = m_B : m_A = 6 : 1$, agreeing with Ceva, and $F$ carries $m_A + m_B = 7$.

Step 3. On $CF$: $CP : PF = m_F : m_C = 7 : 2$, so $m + n = 9$.

Mass points are Ceva in bookkeeping form. If $BD:DC = 3:2$, give $B$ mass $2$ and $C$ mass $3$, so $D$ carries $5$; carrying the masses to the third vertex reads off every remaining ratio without writing a single equation. It works for cevians through one point; for a transversal cutting all three sides, use Menelaus instead.

Ceva ($=1$, cevians concurrent) and Menelaus ($=1$, points collinear) look identical and are distinguished by the configuration, not the formula — so name the configuration before applying either. A quick sanity test: Menelaus always has an odd number of division points OUTSIDE the segments.

Step 1 — masses from $BD:DC = 2:1$: $B = 1$, $C = 2$, so $D = 3$.

Step 2 — from $AE:EC = 3:1$ with $C = 2$: $A\cdot AE = C\cdot EC$ gives $A\cdot3 = 2\cdot1$, so $A = \tfrac23$.

Step 3 — on cevian $AD$, $AP:PD = \text{mass}(D):\text{mass}(A) = 3 : \tfrac23 = 9:2$.

Spotting it. $E$ is not a vertex or a side point but a point ON a cevian; mass points still work, because $E$ is the balance point of $A$ and $D$.

Step 1. $BD : DC = 3 : 2$ gives $m_B = 2$, $m_C = 3$, so $m_D = 5$.

Step 2. $AE : ED = 1 : 2$ puts $E$ nearer $A$, so $A$ is heavier: $m_A \cdot 1 = m_D \cdot 2$, giving $m_A = 10$, and $E$ carries $15$.

Step 3. Line $BE$ passes through the centre of mass, so its other end $F$ balances $A$ and $C$: $AF : FC = m_C : m_A = 3 : 10$. So $m + n = 13$.

Bonus. $F$ carries $13$, so $BE : EF = m_F : m_B = 13 : 2$.

Why it works. Area is $\tfrac12 \cdot \text{base} \cdot \text{height}$, and triangles sharing an apex over one line share the height, so only the bases differ. Mass points supply the ratio in which each cevian is cut at the crossing, and each such ratio splits a triangle into two pieces in that ratio. Chaining two or three such steps reaches any region without a single length or coordinate.

Spotting it. The picture looks the same from every vertex (relabel $A \to B \to C \to A$ and the conditions are unchanged), so one mass-point computation plus symmetry does everything. Name the crossings $X = AD \cap BE$, $Y = BE \cap CF$, $Z = CF \cap AD$.

Step 1. Masses for $AD$ and $BE$: $BD : DC = 1 : 2$ gives $m_B = 2m_C$, and $CE : EA = 1 : 2$ gives $m_C = 2m_A$. Take $m_A = 1$, $m_C = 2$, $m_B = 4$, so $m_D = 6$ and $AX : XD = 6 : 1$.

Step 2. $[ABD] = \tfrac13[ABC]$ because $BD = \tfrac13BC$. Triangle $ABX$ has apex $B$ and base $AX = \tfrac67AD$, so $[ABX] = \tfrac67 \cdot \tfrac13[ABC] = \tfrac27[ABC]$.

Step 3. By the symmetry, $[BCY] = [CAZ] = \tfrac27[ABC]$ as well. These three triangles and the central triangle $XYZ$ fill $ABC$ without overlapping, so $[XYZ] = \left(1 - \tfrac67\right)[ABC] = \tfrac{700}{7} = 100$.

In general (Routh's theorem), when every cevian cuts its side in the ratio $1 : r$ the central triangle is $\tfrac{(r - 1)^2}{r^2 + r + 1}$ of the whole; $r = 2$ gives $\tfrac17$.

Spotting it. The quadrilateral is a triangle ($BCE$) minus a triangle ($BPD$), and both are one or two ratio-steps from the whole.

Step 1. $[BCE] = \tfrac{CE}{CA} \cdot 60 = \tfrac13 \cdot 60 = 20$ (same apex $B$, bases on $CA$).

Step 2. Mass points: $D$ is a midpoint, so $m_B = m_C$; $CE : EA = 1 : 2$ gives $m_C \cdot 1 = m_A \cdot 2$. Take $m_A = 1$, $m_B = m_C = 2$, so $m_D = 4$ and $AP : PD = 4 : 1$.

Step 3. $[ABD] = 30$, and $[BPD] = \tfrac{PD}{AD}[ABD] = \tfrac15 \cdot 30 = 6$.

Step 4. $[PDCE] = [BCE] - [BPD] = 20 - 6 = 14$.