Exploration 9: Pedal Triangles
by Elizabeth Gieseking
A pedal triangle of a point P with respect to is defined as the triangle formed by the feet of the perpendiculars drawn from a point P to the sides of or their extensions. Below, is a pedal triangle of point P with respect to Note that , and
In this figure, point P is outside of and is inside We can also move point P inside as shown below.
If we move point P outside on one of the shorter sides, then only one vertex is on the actual side of and moves outside of
We will now examine some special cases of the pedal triangle.
Case 1: The pedal point P is at the centroid of
The centroid is always inside the triangle. The implication for this on our pedal triangle is that at least two of the three vertices of the pedal triangle will be on the sides of rather than on the extensions of the sides. As a result, at least a portion of the pedal triangle will be on the interior of In the following diagram, only a portion of is on the interior of
A formula has been developed to calculate the area of the pedal triangle of the centroid from the area of the original triangle and its side lengths.
You can download this GSP file to experiment with this relationship.
Case 2: The pedal point P is at the incenter of the triangle.
When the pedal point is located at the incenter of , the resulting pedal triangle will have its vertices at the points of tangency of the incircle with the sides of . Because the incircle is completely within the triangle, the pedal triangle will remain inside , regardless of its shape. This triangle is also called the contact triangle or the intouch triangle. You can download the GSP file to explore this relationship.
If we construct segments , and , we find that they have a point of concurrency called the Gergonne point.
We can use the converse of Ceva’s Theorem to prove that these segments are always concurrent. The line segment joining a vertex of a triangle to any given point on the opposite side is called a cevian. The theorem states: If three cevians , and satisfy , then they are concurrent. We also know that two segments tangent to a circle from the same point are congruent. Therefore and . We can rearrange the equation to get . Thus the three segments are concurrent.
Case 3: The pedal point P is at the orthocenter of the triangle
This case is illustrated below for the cases in which (a) the orthocenter is inside the triangle and (b) the orthocenter is outside the triangle.
We would like to show that when the pedal point is located at the orthocenter of , the pedal triangle is the orthic triangle. The orthic triangle is defined as the triangle whose vertices are the feet of the altitudes of
Because H is the pedal point, , and . Because H is the orthocenter, H is the point of concurrency of the lines containing the altitudes. This implies and are the lines containing the altitudes from A, B, and C, respectively. An altitude is perpendicular to the opposite side or its extension, thus , and . There is only one line through a given point that is perpendicular to a given line, so , and and we conclude that when the pedal point is located at the orthocenter of , the pedal triangle is the orthic triangle of .
Case 4: The pedal point P is at the circumcenter of the triangle
Again we have two subcases as shown below, (a) in which the circumcenter is located on the interior of and (b) in which the circumcenter is located on the exterior of
We wish to show that when the pedal point is located at the circumcenter of , that the pedal triangle is the medial triangle of The medial triangle is defined as the triangle formed by joining the midpoints of the sides of Thus we need to show that C’ is the midpoint of , B’ is the midpoint of and A’ is the midpoint of In the diagram below, we have constructed and and we have replaced the lines from the pedal point with segments to the corresponding sides.
Because O is the circumcenter, it is equidistant from the three vertices of the triangle. Thus This means we have formed three isosceles triangles with base with base and with base Since O is the pedal point, , and . Therefore is the altitude of , is the altitude of , and is the altitude of . An altitude of an isosceles triangle bisects the base, hence C’ is the midpoint of , B’ is the midpoint of and A’ is the midpoint of Thus we can conclude that is the medial triangle of
Case 5: The pedal point P is at the center of the nine point circle for
Just as was true of both the orthocenter and circumcenter, the center of the nine point circle for can lie either on the interior or exterior of This leads to two subcases for the pedal triangle as shown below.
As was true when the pedal point was on the centroid, either two or three of the vertices of the pedal triangle are on the sides of rather than on the extensions. Thus the pedal triangle will never be completely outside of
Case 6: The pedal point is on one of the sides of
When we move the pedal point to one of the sides of the triangle, the intersection of the perpendicular from the pedal point to that side is the pedal point. As seen in the diagram below, when we moved the pedal point to side , A’ was located at the pedal point.
Since A’ is the pedal point, and . Because and are both right angles and since the angles of a quadrilateral sum to 360°, we find that and are supplementary.
Case 7: The pedal point is on one of the vertices of
In this final case, the pedal point is moved to vertex A. As shown below, the pedal triangle has become degenerate and is now a segment from A to A’. Since the lines from the pedal point are perpendicular to the opposite sides, this segment from A to A’ is the altitude from A to side