Question

show that the difference of any 2 sides of a triangle is greater than the third side.

Answer 1

Hey either your question is wrong or incomplete please check and ask again it is possible only on some triangles not every.
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Answer 2

Step-by-step explanation:

let the triangle be ABC

construction

draw the incircle to that triangle with center O

let the incircle intersect the triangle in the points P,Q,R on sides AB,BC,AC respectively

now join OP,OQ and OR

proof

AP = AR = a                                      

BP = BQ = b    

CQ = CR = c

for all the tree equation the reason is : the length of tangents drawn from an external point to two points on a circle are equal

note : a,b,c are value which we assign to them for better calculation

now

take any two sides and subtract them AB - BC = ( AP + BP ) - ( BQ + CQ )         = (a+b) - (b+c) = (a-c)  

AB - BC = (a-c)

AC = CR + AR  = (a+c)

it is known that for any possitive value of a,c    (a-c) is less than (a+c)

so AB - BC < AC

similarly it can be proved for the other sides too

thus it is clear that difference of any 2 sides of a triangle is less than the third side

hope this helps

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