Olympiad Geometry V: Directed Angles, Projective Tools & Named Configurations

This is the discipline that kills configuration dependence. An ordinary angle chase proves the claim for the picture you drew; the moment a point moves to the other side of a line, the ordinary argument needs $180^\circ - \theta$ and the write-up needs a case. Directed angles are immune: lines have no sides, so the one identity covers every configuration at once. The price is small and fixed — you may not talk about "the angle between two segments", only between two LINES, and you may not divide the chase into cases based on position. Graders accept a chase stated in directed angles with no case analysis; they do not accept one that silently assumed a picture.

Ordinary Ceva multiplies side ratios; trig Ceva multiplies sine ratios of the ANGLES the cevians make, which is what you actually control when the cevians are defined by angle conditions (bisectors, isogonals, tangents). It proves in one line that the three symmedians concur, that isogonal cevians of concurrent cevians are concurrent, and that the altitudes, medians and bisectors all concur — three classical theorems, one identity.

The famous pairs are all isogonal conjugates: circumcentre and orthocentre, incenter with itself, centroid and symmedian point, Brocard points with each other. That is why a problem about the symmedian is a problem about the median seen through a mirror — transfer the statement to the median, solve the easy version, reflect back. With $AB = 6$ and $AC = 8$ the symmedian foot splits $BC$ as $36 : 64 = 9 : 16$, while the median splits it $1 : 1$.

A bash is a legitimate solution, not a confession — but the wrong coordinates turn a five-line computation into an hour. The rule of thumb: count the circles. Two or more circles and inscribed angles means complex; ratios along cevians and areas means barycentric; one circle with perpendiculars and explicit lengths means Cartesian, with the circle at the origin. Set up, state what each variable is, and carry the computation to the end — a half-finished bash earns nothing, and graders do read them.

The cross ratio is the only quantity projection preserves, so it is the natural currency of any problem with lots of lines through a point. Four concyclic points have a REAL cross ratio — $1, i, -1, -i$ give $\dfrac{4i}{2i} = 2$ — and the converse holds, which makes it a concyclicity test that survives projection.

Harmonic bundles are self-reproducing: project a harmonic bundle from any point onto any line and it is still harmonic. So one harmonic quadruple found early in a configuration propagates through the whole picture, and statements like "$X$, $Y$, $Z$ are collinear" often reduce to "this bundle is harmonic, and so is that one, so the two lines meet the third at the same point".

With $r = 6$ and $OP = 9$ the polar crosses $OP$ at $OQ = 4$, and the circle meets the line at $\pm 6$: the bundle $(-6, 6; 9, 4)$ has cross ratio $\dfrac{15/3}{10/(-2)} = -1$, harmonic, as every pole–polar pair is. La Hire converts "prove three points are collinear" into "prove three polars concur", which is usually the easier of the two.

A concrete instance: for $A = (0,0)$, $B = (4,0)$, $C = (0,3)$ the point $P = (4,3)$ is the antipode of $A$; its feet on $AB$ and $CA$ are $B$ and $C$ themselves, and the foot on $BC$ is $\left(\tfrac{64}{25}, \tfrac{27}{25}\right)$, which does lie on line $BC$. The "exactly when" makes Simson a collinearity criterion AND a concyclicity criterion, depending on which end of the statement you need.

One circle centred at an arc midpoint carries four points, which is why the lemma is worth its nickname: any problem mentioning both the incenter and the circumcircle almost certainly wants it. The proof is a two-line directed-angle chase — $\angle MBI = \angle MBC + \angle CBI = \tfrac{A}{2} + \tfrac{B}{2} = \angle MIB$, so triangle $MBI$ is isosceles.

In the $13$–$14$–$15$ triangle ($r = 4$, $s = 21$), $\cos^2\tfrac{A}{2} = \dfrac{s(s-a)}{bc} = \dfrac{49}{65}$, so $\rho_A = \dfrac{260}{49}$ — bigger than the incircle, as the picture demands. The radius formula falls out of writing the tangency twice: the centre is on the bisector at distance $\rho/\sin\tfrac{A}{2}$ from $A$, and internal tangency says $OK = R - \rho$.

Let $X$, $Y$, $Z$ be the feet on $BC$, $CA$, $AB$. Since $\angle PZB = \angle PXB = 90^\circ$, the points $P, Z, B, X$ are concyclic; likewise $P, X, C, Y$. Chasing directed angles, $\angle(XZ, XP) = \angle(BZ, BP)$ and $\angle(XP, XY) = \angle(CP, CY)$, so $\angle(XZ, XY) = \angle(BA, BP) + \angle(CP, CA)$. That sum vanishes — i.e. $Z$, $X$, $Y$ are collinear — exactly when $\angle(BA, BP) = \angle(CA, CP)$, which is the concyclicity of $A, B, C, P$. $\blacksquare$ Note how the directed convention made the "iff" free: no case for $P$ on a minor or major arc.

For any cevian $AX$, $\dfrac{BX}{XC} = \dfrac{[ABX]}{[ACX]} = \dfrac{AB\sin\angle BAX}{AC\sin\angle XAC}$. Apply this to the median ($\tfrac{BM}{MC} = 1$) to get $\dfrac{\sin\angle BAM}{\sin\angle MAC} = \dfrac{b}{c}$. The symmedian swaps the two angles, so for it the sine ratio is $\dfrac{c}{b}$ and the side ratio is $\dfrac{c}{b}\cdot\dfrac{c}{b} = \dfrac{c^2}{b^2}$. $\blacksquare$ The same computation with three cevians is exactly trig Ceva, which is why isogonals of concurrent cevians are concurrent.