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Prove that the perimeter of a triangle is greater than the sum of its altitudes.

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Given: ΔABC in which AD⊥BC, BE⊥AC and CF⊥AB

To prove:

AD + BE + CF < AB + BC + AC

Figure:

Proof:

We know that of all the segments that can be drawn to a given line, from a point not lying on it, the perpendicular distance i.e., the perpendicular line segment is the shortest.

Therefore,

Adding (1), (2) and (3), we get

AB+AC+BA+BC+CA+CB>2AD+2BE+2CF

⇒ 2AB+2BC+2CA>2(AD+BE+CF)

⇒ 2(AB+BC+CA)>2(AD+BE+CF)

⇒ AB+BC+CA>AD+BE+CF

⇒ The perimeter of the triangle is greater than the sum of its altitudes

Therefore, Hence proved

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