Question

Hii guyz..

Prove that the difference between any two sides of a triangle is less than its third side.

Solve his question briefly..

All the best!

Answer 1

let assume that a, b, c are the sides of a triangle

Rule- to draw a triangle the addition of two smaller sides should be greater than the third side.( not equal to or not smaller )

if the difference of any two sides of a triangle is not less than its third side, it does not obey the rule…

example

3, 4,7

7–4=3 triangle is not possible

3, 4,8

8–3>4 triangle is not possible

3, 4,6

6–4<3 triangle is possible

Answer 2

Given ---
A ∆ABC

To prove ---
AC - AB < BC

Construction ---
Let AC > AB. Then, take a point D on side AC such that AD = AB. Join BD

Proof ---

AB = AD
We know that angle opposite to equal side is equal
=> ∠1 = ∠2...........(equation 1)

Now,
Side CD of ∆BCD has been produced to A
Therefore, exterior∠ > all other interior ∠s
=> ∠2 > ∠4..........(equation 2)

Similarly,
Side AD of ∆ABD has been produced to C
Therefore, exterior∠ > all other interior ∠s
=> ∠3 > ∠1..........(equation 3)

Now, from equations 1 and 3, we get
∠3 > ∠1 = ∠2
=> ∠3 > ∠2.........(equation 4)

Again, from equations 2 and 4, we get
∠3 > ∠2 > ∠4
=> ∠3 > ∠4
or, ∠4 < ∠3........(equation 5)

Now,
∠4 < ∠3
We know that the side opposite to smaller angle is smaller
=> CD < BC
=> AC - AD < BC.....(since AC = AD + DC)
But AD = AB
Therefore, AC - AB < BC


Similarly we can prove that
BC - AC < AB, and
BC - AB < AC

Hence, it is proved that the difference between any two sides of a triangle is less than its third side.

$proved$
Hope it'll help.. :-D