If I have an angle $\angle{ABC}$, I want to know how to find a ray $\overrightarrow{BD}$ such that $\overrightarrow{BD}$ is in the same plane as $\angle{ABC}$, and the measure of $\angle{ABD}$ is some desired value.

As an example, if I have an angle $\angle{ABC}$ that measures 50 degrees, I want to find a point $D$ such that $\angle{ABD}$ measures 30 degrees, and that $\angle{ABD}$ and $\angle{ABC}$ are on the same plane.

Is there a general formula for this kind of problem?


It seems like you're asking for two separate things at once:

  • An angle that has one ray common with another angle, and whose measure is some fraction of the other angle
  • A point on the new angle that is the same fractional distance between the defining points on the original angle

You can do this if:

  • Point $B$ is the center of a circle on which $A$ and $C$ lie, and
  • Point $D$ also lies on this circle.

In this case, point $D$ will be the same fraction of the way from $B$ to $C$ as the ratio of $\angle ABD$ to $\angle ABC$. In other words, the ratio of the arc lengths is the same as the ratio of the angles.

However, if $B, C, D$ aren't so nicely constrained, in general you won't get both conditions met at once. Absent a circle, the most intuitive way to interpret the distance from $B$ to $C$ would be a straight line, and if $D$ doesn't lie on the line defined by $B$ and $C$, what does "three-fifths of the way from $B$ to $C$" really mean, anyway?

Even constraining $B,C,D$ to be collinear won't get you both conditions at once, because $\angle ABD$ doesn't change proportionally with the distance along $AC$.

  • $\begingroup$ I agree, my question was ambiguous. I removed that ambiguous part, to make clear that I am asking for your first bullet point, not the second one. $\endgroup$ – newprogrammer Jan 16 '15 at 20:54

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