Orthocentre
Orthocentre is the point of concurrence of the altitudes of a triangle. We denote it by H. The triangle obtained by joining the feet of the altitudes is called pedal triangle. The position vector of the orthocentre is given by
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.
