The ratio lemma in olympiad geometry: when to reach for it

Most people meet the ratio lemma once, as a line in someone else's solution, and file it as a trick. That is the wrong file. It is the single most useful bridge in triangle geometry between the two kinds of information a problem gives you: angles at a vertex and lengths on the opposite side. Every time a problem hands you one and asks about the other, this is the tool.

The difficulty is not the proof, which is two applications of the law of sines. It is noticing that the problem is asking you to cross that bridge at all.

The statement

Let $D$ be a point on side $BC$ of triangle $ABC$. Then

$$\frac{BD}{DC} = \frac{AB}{AC}\cdot\frac{\sin\angle BAD}{\sin\angle DAC}.$$

Since $AB/AC = \sin C/\sin B$ by the law of sines, the same thing in angles only is

$$\frac{BD}{DC} = \frac{\sin\angle BAD}{\sin\angle DAC}\cdot\frac{\sin C}{\sin B}.$$

The first form is for when you know sides; the second is for when the problem has given you nothing but angles, which in olympiad geometry is most of the time.

Proof. In triangle $ABD$ the law of sines gives $BD/\sin\angle BAD = AB/\sin\angle ADB$. In triangle $ACD$ it gives $DC/\sin\angle DAC = AC/\sin\angle ADC$. The angles $\angle ADB$ and $\angle ADC$ are supplementary, so their sines are equal. Divide one equation by the other and that common sine cancels.

The same argument works when $D$ is on line $BC$ but outside the segment, because the two angles at $D$ are then equal rather than supplementary, and the sines still match.

The four signals

• A cevian defined by an angle, a question about lengths. "The line through $A$ making $30^\circ$ with $AB$ meets $BC$ at $D$; find $BD/DC$." No similar triangles exist in that figure. The ratio lemma answers it immediately.
• A cevian defined by a length, a question about angles. The converse direction. A median has $BD = DC$, so the lemma says $\sin\angle BAD / \sin\angle DAC = AC/AB$. That is a statement about how the median splits the angle at $A$, and it is exactly what you need when the median is going to be reflected or rotated.
• Three cevians and the word "concurrent" in a problem full of angles. Ceva's theorem wants length ratios. If the problem only gave you angles, apply the lemma three times and multiply: the side lengths cancel in pairs and you are left with the trigonometric form of Ceva.
• Isogonal lines. Any time two lines through $A$ make equal angles with $AB$ and $AC$, the lemma relates their two feet with almost no work, because the sines simply swap.

Watching it work

The angle bisector theorem is the zero case. If $AD$ bisects $\angle A$, the two sines are equal and cancel, leaving $BD/DC = AB/AC$. That is the angle bisector theorem, proved in one line. In the $13$–$14$–$15$ triangle with $AB = 13$, $AC = 15$ and $BC = 14$, the bisector from $A$ lands at $BD = 14\cdot\tfrac{13}{28} = 6.5$ and $DC = 7.5$. The lemma also covers the external bisector, which meets line $BC$ outside the segment in the same ratio $13:15$.

An angle split you cannot do with similar triangles. Take a right isosceles triangle with the right angle at $A$, and a line from $A$ making $30^\circ$ with $AB$ and $60^\circ$ with $AC$. Since $AB = AC$, the lemma gives $BD/DC = \sin 30^\circ/\sin 60^\circ = 1/\sqrt3$. Try finding a pair of similar triangles in that figure that gives the same answer. There is not an obvious one, and that is the point: similar triangles convert lengths to lengths, and the ratio lemma is what you use when the information arrived as an angle.

The symmedian in two lines. Reflect the median $AM$ over the angle bisector to get a line $AK$. By the second signal, the median satisfies $\sin\angle BAM/\sin\angle MAC = AC/AB$. Reflection swaps the two angles, so for $AK$ the sine ratio is $AB/AC$, and the lemma gives

$$\frac{BK}{KC} = \frac{AB}{AC}\cdot\frac{AB}{AC} = \frac{AB^2}{AC^2}.$$

In the $13$–$14$–$15$ triangle that splits $BC$ as $169:225$. Without the lemma, this result usually costs a page.

Ceva, with angles

Multiply three copies of the lemma, one at each vertex, for cevians $AD$, $BE$, $CF$:

$$\frac{BD}{DC}\cdot\frac{CE}{EA}\cdot\frac{AF}{FB} = \frac{\sin\angle BAD}{\sin\angle DAC}\cdot\frac{\sin\angle CBE}{\sin\angle EBA}\cdot\frac{\sin\angle ACF}{\sin\angle FCB}.$$

Every side length appears once on top and once underneath, so they all cancel. By Ceva's theorem the left side equals $1$ exactly when the cevians are concurrent, so the right side does too. That is trigonometric Ceva, and it is nothing more than the ratio lemma used three times.

It makes the classic concurrency results almost free. For the angle bisectors, each sine ratio is $1$, so they meet. For the medians, each length ratio is $1$, so they meet, and the lemma tells you something the length form does not: how the median at each vertex splits that vertex angle. Choose whichever form matches the data the problem gave you.

Where it goes wrong

1. Using the wrong angles. The angles in the lemma are the two pieces of $\angle A$ on either side of the cevian, not the base angles. Write the cevian, then write which angle is on the $B$ side and which is on the $C$ side, before writing any sine.
2. Forgetting the side ratio. $BD/DC = \sin\angle BAD/\sin\angle DAC$ is only true when $AB = AC$. The factor $AB/AC$, or $\sin C/\sin B$, is the part people drop under time pressure.
3. Reaching for it when lengths are all you have. If the problem gives lengths and asks for lengths, the lemma introduces sines you then have to eliminate. Similar triangles or mass points are faster there.

What to do with this

Take five triangle problems that involve a cevian and, for each one, write only one line: what the problem gives you at the vertex, and what it wants on the opposite side. Where one is an angle and the other is a length, write the lemma with the right angles filled in. That line is usually most of the solution.

The Ceva and Menelaus page has the length form and mass points, which handle the all-lengths case. The problem archive has more cevian problems with full solutions, and the daily problem posts a new one every day in three tiers with no account needed.