Q. 154.4( 9 Votes )

# Prove that the pe

Answer :

Given: ∆ABC in which AD perpendicular BC and BE perpendicular AC and CF perpendicular AB.

To prove: AD + BE + CF < AB + BC + AC

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

AD perpendicular BC

AB > AD and AC > AD

AB + AC > 2AD (i)

Similarly,

BE perpendicular AC

BA > BE and BC > BE

BA + BC > 2BE (ii)

And,

CF perpendicular AB

CA > CF and CB > CF

CA + CB > 2CF (iii)

Adding (i), (ii) and (iii), 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.

Hence, proved.

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