Overview

Advanced geometry blends similarity, power of a point, and homothety. Draw accurate diagrams and track equal angles carefully.

Key Ideas

• Ceva's theorem: for concurrent cevians, $\frac{AF}{FB}\cdot\frac{BD}{DC}\cdot\frac{CE}{EA}=1$.
• Homothety maps circles to circles and preserves tangency.
• Power of a point applies to tangents and secants.

Core Skills

Definition. (Cevian) For a triangle $ABC$, a cevian is a line segment from one vertex to a point on the other side of the extension of the aforementioned side. For instance, if $D$ is a point on line $BC$, then $AD$ is a cevian.

Ceva's Theorem

Theorem. (Ceva's Theorem) Let $ABC$ be a triangle. Suppose $D$, $E$, and $F$ are points on $BC$, $CA$, and $AB$, respectively. Then,

if and only if $AD$, $BE$, and $CF$ are concurrent.

Now, we prove this in two parts!

Part I. ($AD$, $BE$, and $CF$ concur $\implies \prod = 1$) Suppose that $AD$, $BE$, and $CF$ concur at $X$. Then,

Multiplying yields the desired result. For more description, $\dfrac{AF}{FB} = \dfrac{[AXC]}{[CXB]}$ holds as each triangle has the same base being $XC$ while the heights are proportional to the ratio $AF:FB$. $\blacksquare$

Part II. ($\prod = 1 \implies$ $AD$, $BE$, and $CF$ concur) Here, we assume $D$, $E$, and $F$ are points on their respective segments such that

Assume that $X = AD \cap BE$ and $CX \cap AB = F'$. Then, from Part I,

Hence, $\dfrac{AF'}{F'B} = \dfrac{AF}{FB}$ which implies $F = F'$, as desired. $\blacksquare$

This is a really important property to not miss: you can use this to prove concurrency, which is a really subtle, but common property in lots of Olympiad Geometry problems, in particular!

Menelaus' Theorem

Theorem. (Menelaus' Theorem) Let $ABC$ be a triangle. Points $P$, $Q$, and $R$ lie on $AB$, $BC$, and $CA$, respectively. Then,

if and only if $AP$, $BQ$, and $CR$ are collinear.

The proof is fairly similar to Ceva's, so it can be an exercise for the reader!

Use Homothety Centers

Identify the homothety center to map tangency points and parallel lines.

Combine with Similarity

Look for similar triangles created by tangents, secants, or parallel lines.

Worked Example

If two tangents from $P$ touch a circle at $A$ and $B$, then $PA=PB$. This follows from equal tangents to a circle.

More Examples

Example 1: Ceva

If $\frac{AF}{FB}=2$, $\frac{BD}{DC}=3$, find $\frac{CE}{EA}$ for concurrency.

$\frac{CE}{EA}=\frac{1}{6}$.

Example 2: Power of a Point

From $P$, a secant meets the circle at $A,B$ with $PA=3$, $PB=12$. Find the tangent length.

$PT^2=36$, so $PT=6$.

Example 3: Homothety

If two circles are externally tangent at $T$, the line joining centers passes through $T$. This is the homothety center.

Strategy Checklist

• Decide if Ceva/Menelaus applies.
• Look for homothety centers to relate segments.
• Use power of a point to connect lengths.

Practice Problems