Prove that the sum of three altitudes of a triangle is less than the perimeter of the triangle.

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Solution

Given: A triangle $A B C$ in which $A D ⊥ B C, B E ⊥ A C$ and $C F ⊥ A B$ .
To prove:
$A D + B E + C F < A B + B C + C A$
or $A D + B E + C F <$ Perimeter of $Δ A B C$
Proof: As we know that from all the segments that can be drawn to a given line, from a point not lying on it, the perpendicular line segment is the shortest one.
$A D ⊥ B C$
$A B > A D$ and $A C > A D$
$A B + A C > 2 A D$ …(i)
$B E ⊥ A C$
$B C > B E$ and $B A > B E$
$B C + B A > 2 A B$ …(ii)
Also $C F ⊥ A B$
$A C > C F$ and $B C > C F$
$A C + B C > 2 C F$ …(iii)
Adding (i), (ii) and (iii), we get
$(A B + A C) + (A B + B C) + (A C + B C) > 2 A D + 2 B E + 2 C F$
$2 (A B + B C + C A) > 2 (A D + B E + C F)$
$A B + B C + C A > A D + B E + C F$
Similarly, Perimeter of $Δ A B C > A D + B E + C F$
Hence proved.

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