I am struggling with finding coordinates of a point for arc origin.

I am programming a tool which converts a corner (intersection of two lines) into an arc of needed radius.


I have intersecting line segments AD and AC.


  • Coordinates of Point A
  • Length of Segment BE (r)
  • Angle between AC and AB (which also equals to angle between AD & AB)

My problem is finding any dimension related to segment AB (length, projections, etc), which if known, I can solve the problem.

Ultimately Need:

  • Points D, B, and C to construct an Arc and resize the intersecting lines to end at respective points C & D.

enter image description here

  • $\begingroup$ What have you tried? $\endgroup$
    – clathratus
    Mar 21 '19 at 18:21
  • $\begingroup$ What do you mean? I am not able to solve it correctly, that's all. I've used AEx projection + rx length to get the new coordinate of rx, but as you can see that would not be correct since it doesn't account for AB segment. I've no idea what to do to get info about the AB segment. $\endgroup$
    – Micard
    Mar 21 '19 at 18:31
  • $\begingroup$ Is $AC$ tangent to the circle at point $C$? if yes, then $AB=r(\frac{1}{\sin \angle{EAC}}-1)$ $\endgroup$
    – Vasya
    Mar 21 '19 at 18:37
  • $\begingroup$ Are AC and AD perpendicular, as shown in the drawing? Or could they meet at some angle other than 90 degrees? $\endgroup$ Mar 21 '19 at 18:46
  • $\begingroup$ @Vasya Thanks, I will try that! $\endgroup$
    – Micard
    Mar 21 '19 at 18:48

$\triangle{ACE}$ is a right triangle, therefore $${r\over AE} = \sin{\angle{CAE}} = \sin{\left(\frac12\angle{CAD}\right)}.$$ This lets you find $E$, and from there $B$, $C$ and $D$ are easily found. Note,too, that $\angle{AEC}=\frac\pi2-\angle{CAE}$.

  • $\begingroup$ Thank you, works pretty well, not sure why I couldn't solve a 7th grade geometry problem... $\endgroup$
    – Micard
    Mar 22 '19 at 13:40

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