Question

Show that the sum of the three altitudes of a triangle is less than the sum of three sides of the triangle. [4 MARKS]

Answer 1

Concept : 1 Mark
Process : 2 Marks
Proof : 1 Mark

Given: A $△$ABC in which AD$\perp$BC, BE$\perp$AC and CF$\perp$AB.

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

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 line segment is the shortest.

Therefore,

AD $\perp$ BC $\Rightarrow$ AB $>$ AD and AC $>$ AD

$\Rightarrow$ AB+AC $>$ AD + AD

$\Rightarrow$ AB+AC $>$ 2 AD .....(i)

BE $\perp$ AC $\Rightarrow$ BC $>$ BE and BA $>$ BE

$\Rightarrow$ BC + BA$>$ BE + BE

$\Rightarrow$ BA + BC $>$ 2 BE ...(ii)

And, CF $\perp$ AB $\Rightarrow$ AC $>$ CF and BC $>$ CF

$\Rightarrow$ AC + BC $>$ 2 CF ...(iii)

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

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

$\Rightarrow$ 2(AB + BC + AC) $>$ 2(AD + BE +CF)

$\Rightarrow$ AD + BE + CF $<$ AB + BC + AC