# Construct triangle ABC from these given data

euclidean-geometrygeometric-constructiongeometry

I want to construct triangle $$ABC$$ and it is given:

• Segment $$BC$$
• $$\angle BAC = \theta$$
• There is a circle through $$B,C$$, the orthocenter $$H$$ of $$\triangle ABC$$ and the centroid $$G$$ of $$\triangle ABC$$.

I noticed that with $$\angle ABC$$, we have that $$\angle BHC = 180^{\circ} – \theta$$ and so we can actually construct the circle that goes through $$BCHG$$. I also thought about the homothety centered at $$G$$ that takes $$H$$ and leads to $$O$$ but I'm not quite sure how to use it.

Here is an approach. I want you to be aware of the variations. The approach remains the same but be aware that positions of orthocenter and circumcenter would change depending on whether $$\theta$$ is acute or obtuse or right and whether the triangle is acute, obtuse or right. For example, if $$\theta$$ is acute, it is not necessary that $$\angle BHC = 180^\circ - \theta~$$. If it is an obtuse triangle, $$\angle BHC = \theta$$. So $$H$$ can either be on the minor arc of the other circle or on the major arc. If $$\theta$$ is obtuse, $$A$$ itself will be on the minor arc of the circumcircle.

The diagram that I have shown assumes $$\theta$$ is acute and for the measurement of $$BC$$ and $$\theta$$ I took, $$\triangle ABC$$ turned out to be an obtuse triangle.

First, to both sides of segment $$BC$$, draw lines at point $$B$$ and $$C$$ measuring angle $$\left(90^\circ - \frac{\theta}{2}\right)$$ to $$BC$$. Say the lines intersect at $$A_1$$ and $$A'$$. Now draw circumcircles to $$\triangle A_1BC$$ and $$\triangle A'BC$$, we call the circles $$S$$ and $$S'$$ respectively and their centers $$O$$ and $$O'$$.

Point $$A$$ that we seek is on one of these circles. If we choose circle $$S$$ for point $$A$$ then $$H$$ and $$G$$ are on $$S'$$.

If $$M$$ is the midpoint of $$BC$$, points $$A_1, O, M, O'$$ and $$A'$$ must be collinear. Also, we know that Euler line divides the circumcenter, centroid and orthocenter $$O, G$$ and $$H$$ in the ratio $$~GH = 2 OG$$. So using power of point $$O$$ with respect to circle $$S'$$,

$$OG \cdot OH = OX \cdot OA' \implies 3 OG^2 = OX \cdot OA'$$

Next, from $$O$$, we draw an arc with length $$OG$$ and it will intersect circle $$S'$$ at two points. We pick one of these points as point $$G$$. We then draw a line through $$M$$ and $$G$$ which is one of the medians. Its intersection with circle $$S$$ gives point $$A$$ and we have $$\triangle ABC$$.

After initial construction, you can look to develop other approaches using a few other relationships, for example we know $$AH = 2 OM$$. I will leave them to you.