Let be the inradius of . The Euler formula gives the distance between the circumcenter and the incenter. Since the circumcenter is on the incircle, , which has the positive solution .

Construction

Draw a circle of radius with center and draw a diameter . Draw a chord at an angle from .

Step 1: Draw a circle with center and radius . Of the two points of intersection of and the segment , let the point be the one closest to .

Step 2: Draw a ray from at an angle from . Let be the perpendicular projection of on . Measure out a point on at distance from .

Step 3: The points and are the intersections of and the line through is perpendicular to .

Verification

Let , and .

Theorem: Let be any triangle. Let be the foot of the altitude from to , and let be the center of the circumscribed circle. Then the angle at between the altitude and equals . The angle between and the angle bisector at is . (See The Plemelj Construction of a Triangle 4.)

Proof of the last part: Let be on the angle bisector at , then .

By construction and the theorem, and is the circumscribed circle of triangle with center and radius .

By construction, is on the angle bisector at and the distance of to is . So the circle with center and radius is the incircle of , which by construction contains .