Problem
GEO-B3-M06-P009 Cosine Form of a Projection
#9
★★☆☆☆ Level 2 of 5
In triangle \(ABC\), prove \(BC=AB\cos B+AC\cos C\).
Inspired by Prasolov trigonometric geometry method
C. Hint 1. Drop the altitude from \(A\) to \(BC\).
D. Hint 2. Split side \(BC\) into two projections.
Let \(D\) be the foot of the altitude from \(A\) to \(BC\). Then \(BD=AB\cos B\) if \(D\) lies on segment \(BC\); in the obtuse case this is interpreted with directed projections.
Similarly, \(DC=AC\cos C\). Adding the projections on line \(BC\), we get \(BC=AB\cos B+AC\cos C\).
This simple equality often replaces a bulky cosine-rule computation.