Let's construct the triangle if the radius of the circumscribed circle, as well as the foot of the altitude, angle bisector, and median from the same vertex are known.

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Suppose that $\triangle ABC$ is consistent with the data, such that the foot of the altitude from vertex $A$ to side $BC$ is $H$, the angle bisector intersects $BC$ at $D$, and the midpoint of $BC$ is $M$. We can further assume that $AB < AC$, and in this case, the order of the points is: $B, H, D, M, C$. Additionally, let's construct the circumcircle of $\triangle ABC$; its diameter passing through $M$ is $EF$, and thus $EF \perp BC$. The point $E$ bisects the arc $\widehat{BAC}$ corresponding to the inscribed angle $\widehat{BEC}$, and therefore the angle bisector $AD$ passes through point $E$. The distances $HM$ and $DM$ are known.

![](https://cdn.mathpix.com/cropped/2024_05_02_2150fc6e91d6309164feg-1.jpg?height=382&width=380&top_left_y=405&top_left_x=859)

Based on this, the construction can be carried out as follows: in a circle with a given radius $R$, we draw one of its diameters, $EF$. We then draw two parallel lines, $e_1$ and $e_2$. Line $e_1$ is drawn at a distance $HM$ from $EF$, and line $e_2$ is drawn at a distance $DM$ from $EF$ (on the same side of $EF$). The former intersects the circle at points $A$ and $A'$. We connect point $A$ (as one of the vertices of the desired triangle) to $E$; $AE$ intersects $e_2$ at point $D$. The chord drawn through $D$ perpendicular to $EF$ will be the side $BC$ of the triangle.

If we connect $A'$ to point $F$, $A'F$ intersects $e_2$ at point $D'$. This results in a triangle that is symmetric (and congruent) to $\triangle ABC$; the axis of symmetry is the diameter perpendicular to $EF$.

The construction is possible if

$$
DM < HM < R
$$

Rajó Sándor (Ref. g. VIII. o. Debrecen.)

Note. $1^{\circ}$. If any two of the points $H, D, M$ coincide, then the third point also coincides with them; in this case, $\triangle ABC$ is isosceles. The problem then becomes indeterminate.

![](https://cdn.mathpix.com/cropped/2024_05_02_2150fc6e91d6309164feg-1.jpg?height=423&width=385&top_left_y=1366&top_left_x=859)

$2^{\circ}$. If we allow the points on side $BC$ to be ordered as: $H, B, D, M, C$ (with $AB < AC$), then a obtuse angle will form at vertex $B$. We get such a triangle if we connect the vertex $A$ (the intersection of the circle and $e_1$) to point $F$; $AF$ intersects $e_2$ at point $D_1$. The chord passing through $D_1$ and perpendicular to $EF$ will be the side $B_1C_1$ of $\triangle AB_1C_1$, where an obtuse angle lies at $B_1$. $AD_1$ bisects $\angle B_1AC_1$; $EF$ bisects side $B_1C_1$, and $B_1C_1$ intersects $e_1$ at point $H_1$. This $H_1$ is the foot of the altitude from vertex $A$. Thus, $\triangle AB_1C_1$ and its symmetric counterpart, $\triangle A'B_1'C_1'$, satisfy the conditions.