Question

prove that sum of three altitudes of TRIANGLE is less than sum of three sides of triangle brainly teachers challenge solve it​

Answer 1

Step-by-step explanation:

Let ∆ABC,have altitudes BE,CF and AD.

Now in ∆BEC,apply Pythagoras theorem

${BC}^{2} = {BE}^{2} + {EC}^{2} \\ or \\ {BE}^{2} = {BC}^{2} - {EC}^{2} \\ thus \\ \\ {BE}^{2} <{BC}^{2} \\ \\ BE < BC \: \: \: ...eq1$

In ∆ADC

${AC}^{2} = {AD}^{2} + {DC}^{2} \\ or \\ {AD}^{2} = {AC}^{2} - {DC}^{2} \\ thus \\ \\ {AD}^{2} <{AC}^{2} \\ \\ AD< AC \: \: \: ...eq2$

now in ∆BCF

${BC}^{2} = {CF}^{2} + {BF}^{2} \\ or \\ {CF}^{2} = {BC}^{2} - {BF}^{2} \\ thus \\ \\ {CF}^{2} < {BC}^{2} \\ \\ CF< BC\: \: \: ...eq3$

from eq1,2 and 3

$AD + BE+ CF < AB + BC + CA$

Thus if is proved that sum of three altitudes of TRIANGLE is less than sum of three sides of triangle.

Hope it helps you.