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In the following diagram, $E$ is a point inside triangle $ABC$, and $\overline{AE}, \overline{BE}, \overline{CE}$ are drawn in. Angles $\alpha_1, \alpha_2, \beta_1, \beta_2, \gamma_1, \gamma_2$ are labeled as shown.

triangle diagram

Considering either the sum of angles of $\triangle ABC$ or the sum of angles around $E$, the angles must satisfy $$ \alpha_1 + \alpha_2 + \beta_1 + \beta_2 + \gamma_1 + \gamma_2 = \pi. \tag{1} $$

But there must be another condition. Indeed, if $\alpha_1, \alpha_2, \beta_1$, and $\beta_2$ are fixed, then $\gamma_1$ and $\gamma_2$ are determined.

So my question is: Given (1), what is the additional relation between the angles $\boldsymbol{\alpha_1}$, $\boldsymbol{\alpha_2}$, $\boldsymbol{\beta_1}$, $\boldsymbol{\beta_2}$, $\boldsymbol{\gamma_1}$, $\boldsymbol{\gamma_2}$?

Somehow I was under the impression that angle chasing suffices to obtain all of the angle dependencies in a diagram. Evidently I am wrong, as angle chasing in the above diagram does not appear to yield anything other than (1).

Caleb Stanford
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1 Answers1

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By the law of sines:

$$ \frac{CE}{BE} = \frac{\sin \alpha_1}{\sin \alpha_2}, \;\;\; \frac{AE}{CE} = \frac{\sin \beta_1}{\sin \beta_2}, \;\;\; \frac{BE}{AE} = \frac{\sin \gamma_1}{\sin \gamma_2} $$

Multiplying together:

$$ \sin \alpha_1\;\sin\beta_1\;\sin \gamma_1 \;=\; \sin \alpha_2\;\sin\beta_2\;\sin \gamma_2 $$

dxiv
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  • Great, thank you. Now this determines, say, $\sin \gamma_1$ and $\sin \gamma_2$ given the rest. But to determine $\gamma_1, \gamma_2$ exactly we need to know if they are between $0$ and $\frac{\pi}{2}$ or between $\frac{\pi}{2}$ and $\pi$. Any comments on this? – Caleb Stanford Aug 25 '16 at 00:29
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    @6005 The above gives an equation $\lambda \sin \gamma_1 = \mu \sin(\theta - \gamma_1)$ where $\lambda, \mu, \theta$ depend on $\alpha_i, \beta_i$ only. That simplifies to $\cot \gamma_1 = \textit{something}$. Since $\cot$ is bijective on $(0, \pi)$ this would have a unique solution $\gamma_1$. – dxiv Aug 25 '16 at 00:59
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    Great, I think that settles it. I will accept after a few days. – Caleb Stanford Aug 25 '16 at 02:00
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    +1. It's worth noting that this is the "trigonometric form" of Ceva's Theorem. – Blue Aug 25 '16 at 05:43
  • @Blue Thank you for the pointer. I was pretty sure there must be some canonical tag to it, but didn't guess the right keywords to find it. – dxiv Aug 25 '16 at 05:53