# Show that the dif

In a triangle let AC > AB

Then, along AC draw AD = AB and join BD

Proof: In Δ ABD,

ABD = 2 (angles opposite to equal sides) ….(ii)

Now, we know that the exterior angle of a triangle is greater than either of its opposite interior angles.

∴∠ 1 >ABD

1 > 2 ….(iii)

Now, from (ii)

2 > 3 ….(iv) (2 is an exterior angle)

Using (iii) and (iv),

1 > 3

BC > DC (side opposite to greater angle is longer)

BC > AC – AB (since, AB = AD)

Hence, the difference of two sides is less than the third side of a triangle

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