Trigonometry/Circles and Triangles/The Pedal Triangle

Given any triangle ABC, let the altitudes through each vertex meet the opposite sides at G, H, K respectively. As already noted, the altitudes intersect at the orthocentre O of the triangle. The triangle GHK is the pedal triangle of ABC. O is its incentre, and ABC is its excentral triangle. If ABC is right-angled, the altitudes all pass through the right angle and the pedal triangle is a single point.

In the quadrilateral BKOG, the angles at K and G are by definition right angles, hence BKOG is cyclic; similarly, so are AHOK and CGOH.

If ABC is not an obtuse triangle and its sides are a, b and c, the angles of GHK are 180º-2A, 180º-2B and 180º-2C and its sides are a.cos(A), b.cos(B) and c.cos(C). If the circumradius of ABC is R, we can write the sides as Rsin(2A), Rsin(2B) and Rsin(2A). If ABC is an obtuse triangle, C being the obtuse angle, the angles of GHK are 2A, 2B and 2C-180º, and the third side is -c.cos(C).

Properties of the pedal triangle
The area of GHK is $1/2$(product of two sides)(sine of included angle) = $1/2$Rsin(2B).Rsin(2C).Rsin(180-2A) = $1/2$R2.sin(2A).sin(2B).sin(2C).

The circumradius of GHK is $${HK \over {2\sin(HGK)}} = {{R\sin(2A)} \over {2\sin(180^0-2A)}} = {R \over 2}.$$

The circumcircle of GHK is the nine point circle of ABC.

The perimeter is 4Rsin(A)sin(B)sin(C). The inradius is 2Rcos(A)cos(B)cos(C) and the three exradii are 2Rcos(A)sin(B)sin(C),  2Rsin(A)cos(B)sin(C) and 2Rsin(A)sin(B)cos(C).


 * $$\frac{OG}{AG} + \frac{OH}{BH} + \frac{OK}{CK} = 1$$.

If the sides are g, h and j, then


 * $$\frac{g}{a^2} + \frac{h}{b^2} + \frac{j}{c^2} = \frac{a^2+b^2+c^2}{2abc}$$.