GeometryDifficulty 65-1000 tagged problems

Trigonometric Ceva

Trigonometric Ceva is the habit of proving cevians concur using ratios of sines of the angles they cut. In contest math, that habit turns a crowded setup into a relation the student can test, bound, count, or compute. MathGrit teaches it as a recognizable signal, a deliberate move, and a final translation back to the original question.

No tagged problems yet We are still curating launch-ready practice for this technique.

The Key Result

$$\frac{\sin\angle BAD}{\sin\angle DAC}\cdot\frac{\sin\angle ACF}{\sin\angle FCB}\cdot\frac{\sin\angle CBE}{\sin\angle EBA} = 1$$

The trigonometric form of Ceva's theorem: three cevians are concurrent exactly when this product of sine ratios of the angles they cut equals one.

Why It Works

1
In $\triangle ABC$ let cevians $AD$, $BE$, $CF$ split the vertex angles. By the Law of Sines in $\triangle ABD$ and $\triangle ACD$, the ratio $\dfrac{BD}{DC}$ can be written as $\dfrac{AB\sin\angle BAD}{AC\sin\angle DAC}$.
2
Writing each of $\dfrac{BD}{DC}, \dfrac{CE}{EA}, \dfrac{AF}{FB}$ this way and multiplying, the side-length factors ($AB, BC, CA$) all cancel in pairs.
3
What remains is exactly $\dfrac{\sin\angle BAD}{\sin\angle DAC}\cdot\dfrac{\sin\angle ACF}{\sin\angle FCB}\cdot\dfrac{\sin\angle CBE}{\sin\angle EBA}$, equal to the ordinary Ceva product $\dfrac{BD}{DC}\cdot\dfrac{CE}{EA}\cdot\dfrac{AF}{FB}$.
4
By ordinary Ceva that product is $1$ iff the cevians concur, so the sine-ratio product is $1$ iff $AD, BE, CF$ are concurrent — handy when the data is angles rather than lengths.

Worked Examples

Example 1

Show, using trigonometric Ceva, that the three internal angle bisectors of any triangle $ABC$ are concurrent (they meet at the incenter).

1
Each bisector splits its vertex angle into two equal halves: $\angle BAD = \angle DAC$, $\angle ACF = \angle FCB$, and $\angle CBE = \angle EBA$.
2
So each sine ratio is a ratio of equal angles: $\dfrac{\sin\angle BAD}{\sin\angle DAC} = 1$, and likewise the other two ratios each equal $1$.
3
Their product is $1\cdot 1\cdot 1 = 1$, which is the trigonometric Ceva condition for concurrency.
4
Therefore the three angle bisectors are concurrent — they meet at the incenter.

Answer:

The product of sine ratios is $1$, so the bisectors concur (at the incenter).

Warm-up

In $\triangle ABC$, two cevians split their vertex angles so that $\dfrac{\sin\angle BAD}{\sin\angle DAC} = \dfrac{3}{2}$ and $\dfrac{\sin\angle ACF}{\sin\angle FCB} = \dfrac{4}{5}$. For all three cevians (the third from $B$) to be concurrent, find $\dfrac{\sin\angle CBE}{\sin\angle EBA}$.

1
Trig Ceva for concurrency: $\dfrac{\sin\angle BAD}{\sin\angle DAC}\cdot\dfrac{\sin\angle ACF}{\sin\angle FCB}\cdot\dfrac{\sin\angle CBE}{\sin\angle EBA} = 1$.
2
Substitute the two known ratios: $\dfrac{3}{2}\cdot\dfrac{4}{5}\cdot\dfrac{\sin\angle CBE}{\sin\angle EBA} = 1$, i.e. $\dfrac{6}{5}\cdot\dfrac{\sin\angle CBE}{\sin\angle EBA} = 1$.
3
Solve: $\dfrac{\sin\angle CBE}{\sin\angle EBA} = \dfrac{5}{6}$.

Answer:

$\dfrac{\sin\angle CBE}{\sin\angle EBA} = \dfrac{5}{6}$

Contest level

Use trigonometric Ceva to prove that the three altitudes of any triangle $ABC$ are concurrent (they meet at the orthocenter).

1
The altitude from $A$ meets $BC$ at $D$ with $\angle ADB = 90^\circ$, so in right triangle $ABD$, $\angle BAD = 90^\circ - \angle B$, and similarly $\angle DAC = 90^\circ - \angle C$. Thus $\dfrac{\sin\angle BAD}{\sin\angle DAC} = \dfrac{\sin(90^\circ - B)}{\sin(90^\circ - C)} = \dfrac{\cos B}{\cos C}$.
2
By the same right-angle argument for the altitude from $C$ (foot $F$ on $AB$): $\dfrac{\sin\angle ACF}{\sin\angle FCB} = \dfrac{\cos A}{\cos B}$; and for the altitude from $B$ (foot $E$ on $CA$): $\dfrac{\sin\angle CBE}{\sin\angle EBA} = \dfrac{\cos C}{\cos A}$.
3
Multiply the three ratios: $\dfrac{\cos B}{\cos C}\cdot\dfrac{\cos A}{\cos B}\cdot\dfrac{\cos C}{\cos A} = 1$ (every cosine cancels).
4
The product is $1$, so by trigonometric Ceva the three altitudes concur.

Answer:

The product of cosine ratios telescopes to $1$, so the altitudes concur (at the orthocenter).

Olympiad / Challenge

Prove the isogonal-conjugate theorem: if cevians $AD$, $BE$, $CF$ of $\triangle ABC$ are concurrent, then their isogonals (each reflected across its vertex's angle bisector) are also concurrent.

1
Concurrency of $AD, BE, CF$ gives, by trig Ceva, $\dfrac{\sin\angle BAD}{\sin\angle DAC}\cdot\dfrac{\sin\angle ACF}{\sin\angle FCB}\cdot\dfrac{\sin\angle CBE}{\sin\angle EBA} = 1$.
2
Reflecting cevian $AD$ over the bisector of $\angle A$ produces the isogonal cevian $AD'$ with $\angle BAD' = \angle DAC$ and $\angle D'AC = \angle BAD$. So its sine ratio is $\dfrac{\sin\angle BAD'}{\sin\angle D'AC} = \dfrac{\sin\angle DAC}{\sin\angle BAD}$ — the RECIPROCAL of the original $A$-ratio.
3
Each vertex behaves the same way, so the trig-Ceva product for the three isogonal cevians is the reciprocal of the original product.
4
Since the original product is $1$, its reciprocal is also $1$, so by trig Ceva the three isogonals concur. (Their common point is the isogonal conjugate of the original point.)

Answer:

The isogonal product is the reciprocal of the original ($= 1$), so the isogonals concur — at the isogonal conjugate.

Going Deeper

Generalization: trig Ceva is the angular twin of ordinary Ceva — it replaces side-segment ratios with sine ratios of the split vertex angles, and the two are interconvertible via the Law of Sines. Taking reciprocals of all three ratios reflects each cevian across its bisector, which is exactly why isogonal conjugates exist (the Olympiad example). The symmedian point, the two isogonic (Fermat) points, and the circumcenter–orthocenter isogonal pairing all flow from this.
Where it appears: reach for trig Ceva when the data is ANGLES rather than lengths — 'three cevians make these angles; are they concurrent?' It cleanly proves the altitudes concur (cosine ratios telescope) and the bisectors concur, and it is the standard tool for symmedian and isogonal-conjugate problems on olympiads.
Pitfall: the sine ratios must be taken in the consistent cyclic order shown — at each vertex, the angle toward the 'previous' side over the angle toward the 'next' side ($\angle BAD$ over $\angle DAC$, then $\angle ACF$ over $\angle FCB$, etc.). Swapping the numerator and denominator at one vertex inverts that factor and breaks the $=1$ test; mismatching the angle labels to the cevian feet is the usual error.

Spot the Signal

Use it when concurrency depends on angles rather than on side ratios.
You can describe the hard part as proving cevians concur using ratios of sines of the angles they cut, but a direct attack starts producing clutter.
The problem rewards preserving structure instead of expanding, listing, or guessing too early.

Learn the Move

Start by writing the product of the three sine ratios and checking it equals $1$.
Name the relation that makes Trigonometric Ceva legal before doing computation.
Use the new relation to replace the messiest part of the problem with a cleaner one.
Translate the result back to the quantity the problem actually asks for.

Avoid These Traps

Using the side-ratio Ceva when only angle information is available.
Applying Trigonometric Ceva because it sounds relevant, without checking the trigger first.
Stopping after spotting the technique instead of finishing the calculation or proof.

MathGrit Coach Note

When the data is angles, switch Ceva from sides to sines.

Try it on:

Prove three cevians are concurrent using the trigonometric form of Ceva.

Back to techniques

GeometryDifficulty 65-1000 tagged problems

Trigonometric Ceva

No tagged problems yet We are still curating launch-ready practice for this technique.

The Key Result

$$\frac{\sin\angle BAD}{\sin\angle DAC}\cdot\frac{\sin\angle ACF}{\sin\angle FCB}\cdot\frac{\sin\angle CBE}{\sin\angle EBA} = 1$$

The trigonometric form of Ceva's theorem: three cevians are concurrent exactly when this product of sine ratios of the angles they cut equals one.

Why It Works

1
In $\triangle ABC$ let cevians $AD$, $BE$, $CF$ split the vertex angles. By the Law of Sines in $\triangle ABD$ and $\triangle ACD$, the ratio $\dfrac{BD}{DC}$ can be written as $\dfrac{AB\sin\angle BAD}{AC\sin\angle DAC}$.
2
Writing each of $\dfrac{BD}{DC}, \dfrac{CE}{EA}, \dfrac{AF}{FB}$ this way and multiplying, the side-length factors ($AB, BC, CA$) all cancel in pairs.
3
What remains is exactly $\dfrac{\sin\angle BAD}{\sin\angle DAC}\cdot\dfrac{\sin\angle ACF}{\sin\angle FCB}\cdot\dfrac{\sin\angle CBE}{\sin\angle EBA}$, equal to the ordinary Ceva product $\dfrac{BD}{DC}\cdot\dfrac{CE}{EA}\cdot\dfrac{AF}{FB}$.
4
By ordinary Ceva that product is $1$ iff the cevians concur, so the sine-ratio product is $1$ iff $AD, BE, CF$ are concurrent — handy when the data is angles rather than lengths.

Worked Examples

Example 1

Show, using trigonometric Ceva, that the three internal angle bisectors of any triangle $ABC$ are concurrent (they meet at the incenter).

1
Each bisector splits its vertex angle into two equal halves: $\angle BAD = \angle DAC$, $\angle ACF = \angle FCB$, and $\angle CBE = \angle EBA$.
2
So each sine ratio is a ratio of equal angles: $\dfrac{\sin\angle BAD}{\sin\angle DAC} = 1$, and likewise the other two ratios each equal $1$.
3
Their product is $1\cdot 1\cdot 1 = 1$, which is the trigonometric Ceva condition for concurrency.
4
Therefore the three angle bisectors are concurrent — they meet at the incenter.

Answer:

The product of sine ratios is $1$, so the bisectors concur (at the incenter).

Warm-up

1
Trig Ceva for concurrency: $\dfrac{\sin\angle BAD}{\sin\angle DAC}\cdot\dfrac{\sin\angle ACF}{\sin\angle FCB}\cdot\dfrac{\sin\angle CBE}{\sin\angle EBA} = 1$.
2
Substitute the two known ratios: $\dfrac{3}{2}\cdot\dfrac{4}{5}\cdot\dfrac{\sin\angle CBE}{\sin\angle EBA} = 1$, i.e. $\dfrac{6}{5}\cdot\dfrac{\sin\angle CBE}{\sin\angle EBA} = 1$.
3
Solve: $\dfrac{\sin\angle CBE}{\sin\angle EBA} = \dfrac{5}{6}$.

Answer:

$\dfrac{\sin\angle CBE}{\sin\angle EBA} = \dfrac{5}{6}$

Contest level

Use trigonometric Ceva to prove that the three altitudes of any triangle $ABC$ are concurrent (they meet at the orthocenter).

1
The altitude from $A$ meets $BC$ at $D$ with $\angle ADB = 90^\circ$, so in right triangle $ABD$, $\angle BAD = 90^\circ - \angle B$, and similarly $\angle DAC = 90^\circ - \angle C$. Thus $\dfrac{\sin\angle BAD}{\sin\angle DAC} = \dfrac{\sin(90^\circ - B)}{\sin(90^\circ - C)} = \dfrac{\cos B}{\cos C}$.
2
By the same right-angle argument for the altitude from $C$ (foot $F$ on $AB$): $\dfrac{\sin\angle ACF}{\sin\angle FCB} = \dfrac{\cos A}{\cos B}$; and for the altitude from $B$ (foot $E$ on $CA$): $\dfrac{\sin\angle CBE}{\sin\angle EBA} = \dfrac{\cos C}{\cos A}$.
3
Multiply the three ratios: $\dfrac{\cos B}{\cos C}\cdot\dfrac{\cos A}{\cos B}\cdot\dfrac{\cos C}{\cos A} = 1$ (every cosine cancels).
4
The product is $1$, so by trigonometric Ceva the three altitudes concur.

Answer:

The product of cosine ratios telescopes to $1$, so the altitudes concur (at the orthocenter).

Olympiad / Challenge

Prove the isogonal-conjugate theorem: if cevians $AD$, $BE$, $CF$ of $\triangle ABC$ are concurrent, then their isogonals (each reflected across its vertex's angle bisector) are also concurrent.

1
Concurrency of $AD, BE, CF$ gives, by trig Ceva, $\dfrac{\sin\angle BAD}{\sin\angle DAC}\cdot\dfrac{\sin\angle ACF}{\sin\angle FCB}\cdot\dfrac{\sin\angle CBE}{\sin\angle EBA} = 1$.
2
Reflecting cevian $AD$ over the bisector of $\angle A$ produces the isogonal cevian $AD'$ with $\angle BAD' = \angle DAC$ and $\angle D'AC = \angle BAD$. So its sine ratio is $\dfrac{\sin\angle BAD'}{\sin\angle D'AC} = \dfrac{\sin\angle DAC}{\sin\angle BAD}$ — the RECIPROCAL of the original $A$-ratio.
3
Each vertex behaves the same way, so the trig-Ceva product for the three isogonal cevians is the reciprocal of the original product.
4
Since the original product is $1$, its reciprocal is also $1$, so by trig Ceva the three isogonals concur. (Their common point is the isogonal conjugate of the original point.)

Answer:

The isogonal product is the reciprocal of the original ($= 1$), so the isogonals concur — at the isogonal conjugate.

Going Deeper

Generalization: trig Ceva is the angular twin of ordinary Ceva — it replaces side-segment ratios with sine ratios of the split vertex angles, and the two are interconvertible via the Law of Sines. Taking reciprocals of all three ratios reflects each cevian across its bisector, which is exactly why isogonal conjugates exist (the Olympiad example). The symmedian point, the two isogonic (Fermat) points, and the circumcenter–orthocenter isogonal pairing all flow from this.
Where it appears: reach for trig Ceva when the data is ANGLES rather than lengths — 'three cevians make these angles; are they concurrent?' It cleanly proves the altitudes concur (cosine ratios telescope) and the bisectors concur, and it is the standard tool for symmedian and isogonal-conjugate problems on olympiads.
Pitfall: the sine ratios must be taken in the consistent cyclic order shown — at each vertex, the angle toward the 'previous' side over the angle toward the 'next' side ($\angle BAD$ over $\angle DAC$, then $\angle ACF$ over $\angle FCB$, etc.). Swapping the numerator and denominator at one vertex inverts that factor and breaks the $=1$ test; mismatching the angle labels to the cevian feet is the usual error.

Spot the Signal

Use it when concurrency depends on angles rather than on side ratios.
You can describe the hard part as proving cevians concur using ratios of sines of the angles they cut, but a direct attack starts producing clutter.
The problem rewards preserving structure instead of expanding, listing, or guessing too early.

Learn the Move

Start by writing the product of the three sine ratios and checking it equals $1$.
Name the relation that makes Trigonometric Ceva legal before doing computation.
Use the new relation to replace the messiest part of the problem with a cleaner one.
Translate the result back to the quantity the problem actually asks for.

Avoid These Traps

Using the side-ratio Ceva when only angle information is available.
Applying Trigonometric Ceva because it sounds relevant, without checking the trigger first.
Stopping after spotting the technique instead of finishing the calculation or proof.

MathGrit Coach Note

When the data is angles, switch Ceva from sides to sines.

Try it on:

Prove three cevians are concurrent using the trigonometric form of Ceva.