I'm looking for an English name of "定比分點公式". There's a page on the Chinese Wikipedia dedicated for the vector version of this formula, but that's only available in Chinese. I copy the relevant section of its content so that you don't have to click the linked Wiki page to see what it's about.

$${\overrightarrow {AD}}={\frac {|{\overrightarrow {CD}}|}{|{\overrightarrow {BC}}|}}{\overrightarrow {AB}}+{\frac {|{\overrightarrow {BD}}|}{|{\overrightarrow {BC}}|}}{\overrightarrow {AC}}$$ wiki median jpg

Here's the first case of the area version of this formula that is well-known in the Chinese literature. It can be found in many Chinese journals and books, say Zhang Jing-zhong's (張景中) book 新概念幾何 (, which literally means "new concept geometry" and which has not been translated into other languages). I take the following picture from Fan's Math World.

$P$ and $Q$ are on the same side of the plane divided by line $AB$, and line segment $PQ$ doesn't not intersect with line $AB$. Point $C$ lies on line segment $PQ$ such that $PC = \lambda PQ$. Then we have $$S_{\triangle ABC} = \lambda S_{\triangle QAB} + (1 - \lambda) S_{\triangle PAB}.$$ fan's math world fixed ratio

  1. $S_{\triangle ABC}$ means "the area of $\triangle ABC$".
  2. I didn't put the word "segment" in between "line $AB$". To understand why, consider the second case of the area version of this formula: line segment $PQ$ intersect with line $AB$ at point $M$. Then $\triangle MAB$ degenerates into line segment $AB$ or $AM$ or $BM$, so it has area zero. However, on the right hand side, the area of both $\triangle PAB$ and $\triangle QAB$ and both coefficients $\lambda$ and $1-\lambda$ are positive. Hence, we have zero LHS and positive RHS, which is absurd.
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Both versions make use of the a point on a line segment whose distance with the endpoints follows a fixed ratio. I asked on Discord Math server, and a guy told me that he would call it something like "convex interpolation", but I didn't find relevant pages on DuckDuckGo the search engine.

  • $\begingroup$ I would call the first vector interpolation $\endgroup$
    – Henry
    Apr 27 at 1:04
  • 1
    $\begingroup$ With your absurd result, it becomes less absurd if you give the areas signs $\endgroup$
    – Henry
    Apr 27 at 1:05
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    $\begingroup$ If you only care about magnitudes, the first result is called Stewart's Theorem $\endgroup$ Apr 27 at 1:07
  • $\begingroup$ ... which has a different Chinese Wikipedia page $\endgroup$
    – Henry
    Apr 27 at 1:09
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    $\begingroup$ the vector denoted $AD$ is a convex combination of $AB, AC.$ That is, $\vec{AD} = p \vec{AB} + q \vec{AC} $ with $p+q=1 \; \; \;$ en.wikipedia.org/wiki/Convex_combination $\endgroup$
    – Will Jagy
    Apr 27 at 1:24

1 Answer 1


The formula is also found in the book Machine Proofs in Geometry by Zhang Jing-zhong (whom you mentioned) coauthored with Chou Shang-Ching and Gao Xiao-Shan, which is available in English (as well as online on the website of one of the authors. However, it is not named in that book either.

The following preliminary result does have a name - it is named the Co-side theorem:

Let $M$ be the intersection of lines $AB$ and $PQ$ and $Q \ne M$. Then we have $\frac{S_{PAB}}{S_{QAB}} = \frac{\overline{PM}}{\overline{QM}}$.

(As with your formula, this also requires the use of signed areas to be correct in all cases.)

The formula $$S_{ABC} = \lambda S_{ABP} + (1-\lambda) S_{ABQ}$$ can be proven easily from the co-side theorem, which is possibly the reason that the authors of Machine Proofs in Geometry only refer to it as "Proposition 1.18" (and later "Proposition 2.9").

When teaching this material I refer to the formula as "linearity of signed area" because, literally, it says that $S_{ABC}$ is an affine-linear function in one of the points $A$, $B$, or $C$. (This also follows from the determinant formula for the area of a triangle.)

  • $\begingroup$ Thank you very much for the link of the book, and your explanation. The introduction of signed area is a great tool that I'll explore. $\endgroup$ Apr 27 at 13:21

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