Question

prove that the sum of all the three altitudes of a triangle is less than the sum of three sides of the triangle

Answer 1

Answer :-

( refer to the attachment )

Applying pythagoras theorem in ∆ADC -

AD² + DC² = AC²

Here, AD² will always be less than AC² because AD² is added to DC² and then we get the value as AC²

So,

AD² < AC²

AD < AC -i

Applying pythagoras theorem in ∆EBC -

EB² + EC² = BC²

Here, EC² will always be less than BC² because EC² is added to EB² and then we get the value as BC²

So,

EC² < BC²

EC < BC -ii

Applying pythagoras theorem in ∆ABF -

AF² + BF² = AB²

Here, BF² will always be less than AB² because BF² is added to AF² and then we get the value as AB²

So,

BF² < AB²

BF < AB -iii

Now, adding the equation i , ii and iii :-

AD + EC + BF < AB + BC + AC

So, sum of altitude of the triangle is less than sum of it's sides.

Hence proved.

Answer 2

Given:

• A triangle ABC in which AD ⊥ BC, BE ⊥ AC and CF ⊥ AB.

To prove:

AD + BE + CF < AB + BC + CA

or AD + BE + CF < less than sum of all sides.

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.

AD ⊥ BC

⇒ AB > AD and AC > AD

⇒ AB + AC > 2AD …(i)

BE ⊥ AC

⇒ BC > BE and BA > BE

⇒ BC + BA > 2AB …(ii)

Also CF ⊥ AB

⇒ AC > CF and BC > CF

⇒ AC + BC > 2CF …(iii)

Adding (i), (ii) and (iii), we get

(AB + AC) + (AB + BC) + (AC + BC) > 2AD + 2BE + 2CF

⇒ 2(AB + BC + CA) > 2(AD + BE + CF)

⇒ AB + BC + CA > AD + BE + CF

⇒ So, sum of altitude of the triangle is less than sum of it's sides.

Hence proved