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


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)


BE perpendicular AC

BA > BE and BC > BE

BA + BC > 2BE (ii)


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.

Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
view all courses

Prove that the anRS Aggarwal & V Aggarwal - Mathematics

If the sides of aRD Sharma - Mathematics

D is any point onNCERT Mathematics Exemplar

In a triangle RD Sharma - Mathematics

In Δ ABCRD Sharma - Mathematics

In Δ PQRRD Sharma - Mathematics

<span lang="EN-USRS Aggarwal & V Aggarwal - Mathematics

Prove that the peRD Sharma - Mathematics

In Fig. 10.25, <iRD Sharma - Mathematics

In a <span lang="RD Sharma - Mathematics