# In acute $\triangle ABC$, show $DE+DF \leq BC$, where $D$, $E$, $F$ are the feet of the altitudes from $A$, $B$, $C$, respectively. [duplicate]

Let $$\triangle ABC$$ be an acute angled triangle. The feet of the altitudes from $$A$$, $$B$$, and $$C$$ are $$D$$, $$E$$, and $$F$$, respectively. Prove that $$DE+DF \leq BC$$ and determine the triangles for which equality holds.

(Note: 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. )

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– Blue
Jan 19, 2019 at 18:21

Let $$F'$$ be a reflection of $$F$$ across $$BC$$, so $$FD = F'D$$. Since $$BC$$ is diameter of circle around cyclic quadrilateral $$BCEF$$ and $$\angle FDA = \angle EDA$$ we see that $$F'$$ is on this circle and that $$F',D,E$$ are colinear. Since $$BC$$ is diameter we have $$BC\geq F'E = F'D+DE = FD+DE$$ • For the first question, look at cyclic quadrilateral $BCEF$ and for the second, yes it is necesery, this means F'D,E are on one chord and it is less then diameter. Jan 21, 2019 at 16:16
Due to the existence of the 9-point circle, the circumradius of the orthic triangle $$DEF$$ equals $$\frac{R}{2}$$.
Since the angles of the orthic triangle equal $$\pi-2A,\pi-2B,\pi-2C$$, your inequality is equivalent to $$\sin(2B)+\sin(2C)\leq 2\sin(A)$$ or to $$2\sin(B)\cos(B)+2\sin(C)\cos(C)\leq 2\sin(B)\cos(C)+2\sin(C)\cos(B)$$ or to $$(\cos B-\cos C)(\sin B-\sin C) \leq 0$$ which is trivial since over the interval $$\left(0,\frac{\pi}{2}\right)$$ the sine function is increasing and the cosine function is decreasing. Equality holds only for isosceles triangles.