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About Brocard point in a triangle

Mih´aly Bencze 1

In this paper we present some geometrical inequalities related to the Brocard point in a
triangle.

Definition 1. In any triangle ABC there exists a unique point P for each ∢PAB = ∢PBC =
∢PCA = ω. The point P is the Brocard point, they are named after Henri Brocard (1845-1922),
a French mathematician.

This point P is called the first Brocard point, and ω the Brocard angle.
Theorem 1. (Brocard). In any triangle ABC holds the identity:

ctg ω = ctg A + ctg B + ctg C.
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Proof. The given relation is equivalent to sin(A − ω) · sin(B − ω) · sin(C − ω) = sin ω. But we
sin(A − ω) CP sin(B − ω) AP sin(C − ω) BP
have = , = , = and after multiplication holds
sin ω AP sin ω BP sin ω CP
the identity. □
Remark 1. We have the following relations

P 2 Q
a 1 + cos A
X
ctg ω = ctg A = = Q
4sr sin A
P 2 P
sin A a sin A
= Q = P ,
2 sin A a cos A
2
2
2
2
csc ω = csc A + csc B + csc C,
2sr
.
sin ω = pP
2 2
a b
Corollary 1. Denote P the Brocard point in the triangle ABC, then
AP BP CP
+ + = 2 cos ω.
c a b
Proof. In the triangle APC we have ∢CAP = A − ω, ∢CPA = π − A and using the sine rule
we can write
sin(A − ω) sin A
= .
CP b
Similarly hold
sin(B − ω) sin B sin(C − ω) sin C
= , = .
AP c PB a
Expanding the expressions sin(A − ω), sin(B − ω), sin(C − ω) we get

CP
= cos ω − sin ωctg A,
b
1
Profesor dr., Bras , ov, benczemihaly@gmail.com

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