3.3 Isosceles Triangles, Midsegments & Points of Concurrency

Isosceles and Equilateral Triangles

An isosceles triangle has two congruent sides, called legs; the third side is the base. The Isosceles Triangle Theorem states that the base angles, the angles opposite the congruent legs, are congruent. Its converse is equally testable: if two angles of a triangle are congruent, then the sides opposite them are congruent, so the triangle is isosceles. The angle between the two legs is the vertex angle.

Because a triangle's interior angles sum to 180 degrees, isosceles problems are usually one quick equation. If the vertex angle is V, each base angle equals (180 - V)/2. If a base angle is B, the vertex angle equals 180 - 2B. An equilateral triangle is the special case in which all three sides are congruent; by the theorem it is also equiangular, so every angle measures 60 degrees.

A powerful structural fact: in an isosceles triangle, the segment from the vertex angle to the midpoint of the base is simultaneously the median, the altitude, the angle bisector of the vertex, and the perpendicular bisector of the base. This coincidence is why so many isosceles proofs work, because that one segment hands you a right angle, a bisected base, and a bisected vertex angle all at once.

Worked example. In isosceles triangle RST, legs RS and RT are congruent with RS = x + 5 and RT = 2x - 3. Setting the legs equal gives x + 5 = 2x - 3, so x = 8 and each leg is 13. If the vertex angle at R were 40 degrees, each base angle would be (180 - 40)/2 = 70 degrees.

The Triangle Midsegment Theorem

A midsegment joins the midpoints of two sides of a triangle. The Triangle Midsegment Theorem makes two claims at once: the midsegment is parallel to the third side, and its length is exactly half of that third side. So if the third side is 18, the midsegment is 9; conversely, a midsegment of length 7 is parallel to a side of length 14. Each triangle has three midsegments, and together they split the triangle into four congruent smaller triangles. Do not confuse this with the trapezoid midsegment in Section 3.4, which equals the average of the two parallel bases rather than half of one side.

Points of Concurrency

When three or more lines pass through a single point, they are concurrent. Every triangle has four classic points of concurrency, each built from a different set of special segments. NY Next Gen standard GEO-G.CO.10 explicitly requires the median-to-centroid result, and the perpendicular-bisector and angle-bisector points follow from the locus theorems in GEO-G.CO.9.

Two of these deserve extra attention because they are the most exam-relevant:

• The centroid divides each median so the distance from the vertex to the centroid is twice the distance from the centroid to the midpoint (a 2:1 ratio). If a median is 12 units long, the centroid lies 8 units from the vertex and 4 units from the midpoint.
• The circumcenter lies on all three perpendicular bisectors because every point on a segment's perpendicular bisector is equidistant from its endpoints (the locus theorem). Being equidistant from all three vertices, it is the center of the circle through the vertices. The incenter, on the angle bisectors, is equidistant from the three sides and is the center of the circle inside the triangle.

Location Facts Worth Knowing

In a right triangle, the circumcenter falls on the midpoint of the hypotenuse, and the orthocenter sits at the right-angle vertex. The incenter and centroid always fall inside the triangle, while the circumcenter and orthocenter can fall outside it (in an obtuse triangle). A construction-style item may ask you to locate a point equidistant from the two sides of an angle and from two points A and B; the answer is the intersection of the angle bisector (incenter-style locus) with the perpendicular bisector of AB (circumcenter-style locus).

A Quick Concurrency Application

Suppose the medians of a triangle meet at centroid G, and one median runs from vertex A to midpoint M of the opposite side with AM = 15. Then AG is two-thirds of 15, or 10, and GM is one-third, or 5, because the vertex piece is always the longer 2-part portion. This 2:1 split is a favorite Part II setup and is often paired with the fact that a single median divides a triangle into two smaller triangles of equal area.

Common Traps

• Halving versus averaging: the triangle midsegment is half the third side, while the trapezoid midsegment is the average of two bases.
• Reversing the centroid ratio; the longer 2-part piece is on the vertex side.
• Swapping the circumcenter (equidistant from vertices) with the incenter (equidistant from sides).