The coordinates of $A=(x_{0},y_{0}$) and $B=(x_{1},y_{1}$) are given. How to find the coordinates of $C$ and $D$ as per given information below.

  1. ABC is equilateral triangle such that $AB=BC=CA$
  2. $\angle ADB=120^\circ$
  3. CD line is the bisector line to $\angle ADB$, such that $\angle ADC=\angle CDB=60^\circ$
  4. The points $A,B,C,D$ all are in the same circle

enter image description here

  • $\begingroup$ Setting up (1) easiest part, depending on your method of construction. Construct the equilateral triangle $ABC$ by constructing the circles centred at $A$ and $B$, respectively, with radius $AB$. Pick one of the points of intersection and call it $C$. Connect them for your triangle. From there, it is on you to make the circle $ABC$ using a similar sort of trick. $\endgroup$ – Terra Hyde Jul 15 '15 at 3:42
  • $\begingroup$ I just noticed you want the coordinates of the points. If you are allowed to use trigonometry, this becomes really easy. You could do it algebraically, but it gets a bit messy. $\endgroup$ – Terra Hyde Jul 15 '15 at 3:49
  • $\begingroup$ C lives on the perpendicular bisector of AB, and we know how far it is from AB (at least in terms of the given coordinates). Once the circle is drawn, D is a red herring in that any point on the circular arc between A and B will work. $\endgroup$ – lulu Jul 15 '15 at 3:59

It is not possible to fix the coordinates of $D$ with the given information. In fact, $D$ could be any point on the small arc $\newcommand{arc}[1]{\stackrel{\Large\frown}{#1}}\arc{AB}$.

Given points $A$ and $B$, finding points $C$ and $O$ are not hard. But let's look at the requirements for $D$. Requirements (1) and (4) are met if we choose any point in the small arc $AB$.

The arcs $AC$ and $BC$ clearly have a central angle of $120°$ each, so the large arc $ACB$ has the central angle $240°$. Any angle on the circle that subtends that arc will have half that measure, $120°$. That satisfies requirement (2).

Then angle $\angle ADC$ subtends an angle of $120°$ and thus has a measure of $60°$, and the same is true for $\angle CDB$. That satisfies requirement (3).


By an early Euclid's theorem property of circles the angle subtended at periphery D is half that subtended at center O. Angles

$$ CDB = \frac12 \cdot COB,ADC = \frac12 \cdot AOC $$

In other words all the bisectors at D meet at a single focal concurring point C.

You have picked a particular instance from general situation, so there can be no unique solution or defining position for point D. Meaning, all points on minor arc AB equally well satisfy as solution position of D for stated conditions.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.