Show that the difference of any two sides in a triangle is smaller than the third side.

Explanation

Given : `DeltaABC` in which `AC gt AB`.
To prove : `AC-AB lt BC`
Construction : Cut AB = AD from AC. Join BD. ltbr. Proof : `angleBDC` is the exterior angle of `DeltaABD`.
`:. angle BDC-angleABD+ angleBAD`.
( `:'` the exterior angle is equal t the sum of opposite interior angles of a triangle.)
`implies" "angleBDC gt angleABD` ...(1)
Again `angleADB` is the exterior angle of `DeltaDBC`
`:. angle ADB=angleDBC+angleBCD`
`implies" "angleADB gt angle DBC`
but `AB = AD`
`:. angle ABD = angleADB`
Therefore `n=angleABD gt angle DBC` ...(2)
From (1) and (2), we have
`angleBDC lt angleDBC`
`implies" "BC gt DC" "( :'" side opposite to larger angle is larger")`
`implies" "DC lt BC`
`implies" "AC-AD lt BC`
`implies" "AC-AB lt BC" "( :' AD=AB)`