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.