# BM01 2011/12 Question 6 Geometry Problem

1. Let $ABC$ be an acute-angled triangle. The feet of the altitudes from $A,B$ and $C$ are $D, E$ and $F$ respectively. Prove that $DE +DF \le BC$ and determine the triangles for which equality holds. The altitude from A is the line through A which is perpendicular to BC. The foot of this altitude is the point D where it meets BC. The other altitudes are similarly defined.

Thanks in advance for any contributions.

• Did you mean $DE +DF \ge BC$? – Stefan4024 Mar 9 '15 at 19:27

$\Delta DEF$ is the orthic triangle of $\Delta ABC$, and it is well-known that $DF=b\cos B$ and $DE=c\cos C$. Also using the well-known projection formulae, $a=b\cos C+c\cos B$. So the inequality is equivalent to: $$b\cos B + c\cos C \le b\cos C+c\cos B$$ $$(b-c)(\cos B - \cos C) \le 0$$

This inequality is easily proved because $b\le c \implies \sin B \le \sin C \implies \cos B \ge \cos C$. Similar procedure applies for the $b \ge c$ case.

The equality clearly holds in isosceles triangles. $A,B,C$ and $a,b,c$ have usual meaning.