Orthocentre
The perpendiculars from the vertices to the opposite sides are called the altitudes and they are concurrent at the orthocentre H. In the case of an acute-angled triangle the orthocentre is inside it, but in the case of a right-angled triangle the orthocentre coincides with the right angled vertex, and for an obtuse-angled triangle the orthocentre is outside, in the area above the obtuse vertex. Each of the points A, B, C, H is in fact the orthocentre of the triangle formed by the other three. In any such quadruplet just one of the triangles is acute-angled.
The angles between the altitudes, at H, are the same as the angles of the triangle ABC. (Since, for example CF is perpendicular to AB and BE is perpendicular to AC, it follows that CF, BE are at the same angle to each other as AB, AC.) The triangles AEF, BFD, CDE are thus all similar to ABC.
The triangle DEF is called the orthic triangle (or pedal triangle) of ABC. In the acute-angled case it is the inscribed triangle of least perimeter. If the sides are mirrors (or the cushions of a billiard table) a ray of light (or billiard ball) directed along DE will be reflected to follow the path DEFDEF... The angles of incidence and reflection are equal: AEF = CED, BDF = CDE, BFD = AFE. The sides of this triangle are of lengths a.cosA, b.cosB, c.cosC.
The area of a triangle is half base times height. Let AD = h and DB = x then by Pythagoras: c² = h² + x² and b² = h² + (a–x)², whence, eliminating x we find: h = √(2.a².b² + 2.b².c² + 2.c².a² − a4 − b4 − c4)/2a. Thus the area D = √(2.a².b² + 2.b².c² + 2.c².a² − a4 − b4 − c4)/4. Alternatively, eliminating h we find: x = (a² + c² − b²)/2a.
The orthocentre H of ABC is the incentre of the orthic triangle DEF. In fact the sides AB, BC, CA are the bisectors of the 'external angles' of the triangle DEF and A, B, C are the centres of the three excircles that touch all three sides of DEF externally.
