Remember, altitudes of a triangle are perpendicular lines to the opposite side of the vertex from where they are drawn.
Example Solution:
Method: See solution.
Practice makes perfect
If A, B and O are noncollinear, they cannot all fall on the same line. Also, the orthocenter of a triangle is the point of concurrence for the triangle's altitudes.
Let's connect A and B, which will form our first side.
Since the altitude from a vertex is perpendicular to the line containing the opposite side, we know that the last vertex, C, must fall on the line through O that is perpendicular to AB. To draw this line, we open a compass to a width that is greater then the distance between O and AB and draw an arc that intersects AB twice.
Next, use the compass to draw two more arcs below O, using the first arc's points of intersection with AB as centers.
The line that contains O and the arc's point of intersection is the perpendicular line to AB.
Next, we will draw a ray from B and through O, which will be the altitude drawn from B.
In order for BO to be the altitude of B, we have to draw AC so that it is perpendicular to BO. Therefore, using A as the center, draw an arc that intersects BO twice.
Next, with the same compass setting, draw a pair of intersecting arcs above BO, using the intersection points of the first arc and BO as centers.
The ray that contains A and the intersection of the arcs is perpendicular to BO. The point where it intersects the first altitude we drew is where we find the location of C.
We complete the triangle by connecting the final two sides. Notice that since at least two altitudes of the triangle goes through O, the altitude from A will automatically do so as well.
