1 Example G1: Triangles with circumcenter on a median. Prove that if the circumcenter of a triangle lies on a median, the triangle either is isosceles or contains a right angle. D D
2 Solution to Example G1: Triangles with circumcenter on a median. Prove that if the circumcenter of a triangle lies on a median, the triangle either is isosceles or contains a right angle. D D Solution. Suppose the circumcenter of triangle lies on the median D. (i) If D =, then D = D = D. The triangles D and D are isosceles, and = D and = D. Therefore, = D + D = + = (180 ). From this, =90. (ii) If and D are distinct points, the triangles D and D are congruent by the SSS test. Therefore, D = D =90 and the median D (containing D) is perpendicular to. From this, triangles D and D are congruent by the SS test. It follows that =.
3 The orthocenter The three altitudes of a triangle are concurrent at the orthocenter of the triangle. Proof. Given triangle, construct the parallels through the vertices to their opposite sides. These bound a triangle (the superior triangle). and are both parallelograms. = =. is the midpoint of. The altitude of through is perpendicular to. Therefore, it is the perpendicular bisector of. Similarly, the altitudes of through and are the perpendicular bisectors of and respectively. Therefore, these three altitudes are concurrent at a point, which is the circumcenter of (the superior) triangle. This is called the orthocenter of triangle.
4 Example G3. Triangle has orthocenter. Prove that triangle has orthocenter. Similarly, triangle has orthocenter, and triangle has orthocenter.
5 Solution to Example G3. Triangle has orthocenter. Prove that triangle has orthocenter. Similarly, triangle has orthocenter, and triangle has orthocenter. Proof. The orthocenter of a triangle is the intersection of two of its altitudes. Let be the orthocenter of. Since and, and are two altitudes of triangle. Their intersection is the orthocenter of.
6 Example G4: construction of triangle. onstruct triangle given the vertices, and the orthocenter. Solution. (1) onstruct the perpendicular from to the line. (2) onstruct the perpendicular from to the line. (3) The intersection of the two lines in (1) and (2) is the vertex.
7 Example G2: The orthocenter. Given triangle with circumcenter, orthocenter, and midpoint D of, prove that =2 D. D
8 Example G3: The orthocenter lies on the line G. G The superior triangle is the image of under the homothety h(g, 2): G = 2 G, G = 2 G, G = 2 G. The orthocenter, being the circumcenter of, is the image of the circumcenter of under the same homothety. G = 2 G. The line containing, G, and is called the Euler line of triangle.
9 Example G4: Reflections of the circumcenter. The reflections of the circumcenter in the sidelines of triangle form a triangle oppositely congruent to at the midpoint of. N If a is the reflection of in, then a =2 D =. The quadrilateral a has two sides equal and parallel. It is a parallelogram. The diagonals a and bisect each other. Therefore, a and has the same midpoint N. a Similarly, the midpoint N of is also the midpoints of b and c, where b, c are the reflections of in and respectively. This means that a b c and are oppositely congruent at : N a = N, N b = N, N c = N.
10 Example G4: Reflections of the circumcenter. The reflections of the circumcenter in the sidelines of triangle form a triangle oppositely congruent to at the midpoint of. N Since a is a parallelogram, a =. Similarly, b =, c =. Therefore, is the circumcenter of a b c, and the radius of the cirumcircle is R. a
Example G5: The reflections of the orthocenter. The reflections of the orthocenter in the sidelines lie on the circumcircle. 11 a Let a be the reflection of in. Since a and a are the reflections of and in, the quadrilateral a a is a trapezoid symmetric in. Its diagonals are equal in length. Since a = R, we also have a = R. The reflection of in lies on the circumcircle. Similarly, the reflections of in and also lie on the circumcircle. a