Geometry: Coordinates & Complex

A polygon with vertices $(x_i,y_i)$ in order has area $\tfrac12\left|\sum_i (x_iy_{i+1} - x_{i+1}y_i)\right|$.

In plain terms. List the corner coordinates in a loop, cross-multiply neighbors, and half the total gives the area — no need to cut the shape up.

Example. Triangle $(0,0),(4,0),(0,3)$: area $= \tfrac12|0\cdot0 - 4\cdot0 + 4\cdot3 - 0\cdot0 + 0 - 0| = 6$.

Cevians $AD$, $BE$, $CF$ (with $D, E, F$ on sides $BC$, $CA$, $AB$) are concurrent if and only if $\dfrac{BD}{DC}\cdot\dfrac{CE}{EA}\cdot\dfrac{AF}{FB} = 1$.

In plain terms. Three lines from the corners of a triangle to the opposite sides pass through one common point exactly when the three side-splitting ratios multiply to $1$. Two known ratios always lock in the third — no lengths of the triangle itself are ever needed.

Example. If $BD:DC = 2:3$ and $CE:EA = 1:2$ and the cevians are concurrent, then $\tfrac23\cdot\tfrac12\cdot\tfrac{AF}{FB} = 1$ forces $AF:FB = 3:1$.

A transversal line crossing lines $BC$, $CA$, $AB$ of triangle $ABC$ at $D$, $E$, $F$ satisfies $\dfrac{BD}{DC}\cdot\dfrac{CE}{EA}\cdot\dfrac{AF}{FB} = 1$ (unsigned; with signed ratios the product is $-1$).

In plain terms. Same product as Ceva, but for a straight LINE slicing across the triangle instead of three concurrent cevians — the line always cuts one side externally, on its extension. Use it whenever a problem draws a line through two marked points and asks where it crosses a third side.

Example. With $BD:DC = 2:3$ on $BC$ and $M$ the midpoint of $CA$ ($CM:MA = 1$), the line $DM$ meets line $AB$ at $X$ with $\tfrac{AX}{XB}\cdot\tfrac23\cdot 1 = 1$, so $AX:XB = 3:2$ — and $X$ lies outside segment $AB$.