It is possible to find distance of orthocentre from vertices and sides of a triangle.

Let H be the orthocentre

In the figure, we observe that

HK = BK tan ∟HBK = BK tan ∟LBC

HK = BK tan (90°-C) (in ΔBHK)

= AB cos B cot C

=c cos B cos C/ sin C = (c/sin C) cos B cos C

= 2 R cos B cos C

Similarly, we can prove that

HL = 2R cos A cos C

and       HM = 2R cos A cos B

Now, AH = AL sec (90 – C)

= AB cos A cosec C

= (c/sin C) cos A = 2 R cos A

Similarly we get BH = 2R cos B, CH = 2R cos C

Thus, distances of orthocentre from sides a, b, c, are given by
2 R cos B cos C, 2R cos A cos C, 2 R cos A cos B respectively and from vertices A,B,C are given by 2R cos A, 2R cos B and 2R cos C  respectively.

ΔKLM is pedal triangle for ΔABC

In quadrilateral BKHM,

∟BMH +∟ BKH= 90° + 90° = 180°

Þ         BKHM is a cyclic quadrilateral

Þ         ∟BMK = ∟BHK = 90° – ∟HBK

= 90° – ∟LBC

= 90° – (90° – C) = C

=> ∟KMH = 90° – C

Similarly, we can prove that

∟HML = 90° – C

=> ∟KML = 180° – 2 C

Similarly

∟KLM = 180° – 2B and ∟MKL = 180° – 2A

i.e. angles of pedal triangle are supplement of double of opposite angles of the original triangle

Now, in Δ AML,

Similarly, it can be shown that KM = b cos B, KL = c cos C

i.e. sides of pedal triangle are a cos A, b cos B, c cos C.

Illustration

1. MKL is the pedal triangle of ABC; prove that its area is

2 S cos A cos B cos C, where S is area of ΔABC.

Circumcircle is a circle circumscribing the triangle i.e. passing through the three vertices of the triangle. Its center is naturally found to be at the point of intersection of perpendicular bisectors of the sides of the triangle. Its centre is usually denoted by O and it’s radius by R and called circumradius. It may happen that the circumcentre is lying out side the triangle as any angle may be out use. If O’ denotes the origin, then the position vector of circumcentre O is

Incircle

A circle touching all the three sides of a triangle is said to be incircle. It’s centre lies at the point of intersection of angle bisectors of the angles of the triangle. It’s centre is denoted by I and radius by r and called inradius. This is a circle inscribed in a triangle. Its centre can never lie outside the triangle. Its position vector is given by

Escribed circle

It is a circle touching on of the three sided externally and the rest two extended sides internally. Obviously, every triangle will lead to three escribed circles. The circle opposite to vertex A gets it’s centre denoted by I₁ and radius by r₁ called exradius. Similarly, circles opposite to vertex B and C get their centers and radii denoted by I₂, r₂ and I₃, r₃ respectively. Obviously its centre will lie on the point of concurrence of two external angle bisectors and one internal angle bisector. The position vector of the escribed centre I₁ is given by

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