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. A A O B D C B D C 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. A A O B D C B D C Solution. Suppose the circumcenter O of triangle ABC lies on the median AD. (i) If D = O, then AD = BD = CD. The triangles ADB and ADC are isosceles, and ∠B = ∠BAD and ∠C = ∠CAD. Therefore, A = ∠BAD + ∠CAD = B + C = (180◦ − A). From this, A =90◦. (ii) If O and D are distinct points, the triangles DOB and DOC are congruent by the SSS test. Therefore, ∠ODB = ∠ODC =90◦ and the median AD (containing OD) is perpendicular to BC. From this, triangles ADB and ADC are congruent by the SAS test. It follows that AB = AC. 3 The orthocenter The three altitudes of a triangle are concurrent at the orthocenter of the triangle. C A B H B C A Proof. Given triangle ABC, construct the parallels through the vertices to their opposite sides. These bound a triangle ABC (the superior triangle). ABCB and ACBC are both parallelograms. AB = BC = AC. A is the midpoint of BC. The altitude of ABC through A is perpendicular to BC. Therefore, it is the perpendicular bisector of BC. Similarly, the altitudes of ABC through B and C are the perpendicular bisectors of CA and AB respectively. Therefore, these three altitudes are concurrent at a point H, which is the circumcenter of (the superior) triangle ABC. This is called the orthocenter of triangle ABC. 4 Example G3. Triangle ABC has orthocenter H. Prove that triangle BCH has orthocenter A. Similarly, triangle CAH has orthocenter B, and triangle ABH has orthocenter C. 5 Solution to Example G3. Triangle ABC has orthocenter H. Prove that triangle BCH has orthocenter A. Similarly, triangle CAH has orthocenter B, and triangle ABH has orthocenter C. A H B C Proof. The orthocenter of a triangle is the intersection of two of its altitudes. Let H be the orthocenter of ABC. Since BH ⊥ AC and CH ⊥ AB, AC and AB are two altitudes of triangle BCH. Their intersection A is the orthocenter of BCH. 6 Example G4: construction of triangle. Construct triangle ABC given the vertices B, C and the orthocenter H. Solution. (1) Construct the perpendicular from B to the line CH. (2) Construct the perpendicular from C to the line BH. (3) The intersection of the two lines in (1) and (2) is the vertex A. A H B C 7 Example G2: The orthocenter. Given triangle ABC with circumcenter O, orthocenter H, and midpoint D of BC, prove that AH =2· OD. C A B O H B D C A 8 Example G3: The orthocenter lies on the line OG. C A B O G H B C A The superior triangle ABC is the image of ABC under the homothety h(G, −2): GA = −2 · GA, GB = −2 · GB, GC = −2 · GC. The orthocenter H, being the circumcenter of ABC, is the image of the circumcenter O of ABC under the same homothety. GH = −2 · GO. The line containing O, G, and H is called the Euler line of triangle ABC. 9 Example G4: Reflections of the circumcenter. The reflections of the circumcenter O in the sidelines of triangle ABC form a triangle oppositely congruent to ABC at the midpoint of OH. A O N H B C Oa If Oa is the reflection of O in BC, then OOa =2· OD = AH. The quadrilateral AHOaO has two sides equal and parallel. It is a parallelogram. The diagonals AOa and OH bisect each other. Therefore, AOa and OH has the same midpoint N. Similarly, the midpoint N of OH is also the midpoints of BOb and COc, where Ob, Oc are the reflections of O in CA and AB respectively. This means that OaObOc and ABC are oppositely congruent at B: NOa = −NA, NOb = −NB, NOc = −NC. 10 Example G4: Reflections of the circumcenter. The reflections of the circumcenter O in the sidelines of triangle ABC form a triangle oppositely congruent to ABC at the midpoint of OH. A O N H B C Oa Since AHOaO is a parallelogram, HOa = AO. Similarly, HOb = BO, HOc = CO. Therefore, H is the circumcenter of OaObOc, and the radius of the cirumcircle is R. 11 Example G5: The reflections of the orthocenter. The reflections of the orthocenter in the sidelines lie on the circumcircle. A O H B C Ha Oa Let Ha be the reflection of H in BC. Since Oa and Ha are the reflections of O and H in BC, the quadrilateral OOaHaH is a trapezoid symmetric in BC. Its diagonals are equal in length. Since HOa = R, we also have OHa = R. The reflection of H in BC lies on the circumcircle. Similarly, the reflections of H in CA and AB also lie on the circumcircle..