Question

prove that the difference of any two sedes of a triangle is less than the third side​

Answer 1

Step-by-step explanation:

To prove: AC-AB< BC

Construction: Mark a point 'D' on AC such that the distance AB=AD

Therefore, we have to prove that AC-AD<BC

So, we have to prove that CD < BC

Proof:

Consider triangle ABD,

since AB=AD

Since, opposite angles opposite to the equal opposite sides are always equal.

So,

So, p = p

Now, let

Now, using exterior angle property in triangle ABD,

Exterior angle property of a triangle states that the measure of an exterior angle is equal to the sum of the two interior angles.

(equation 1)

Now, using exterior angle property in triangle BCD,

(equation 2)

Now, by comparing equation 1 and 2,

So, y > x

BC > CD

CD < BC

AC - AD < BC

Since, AD = AB

Therefore, AC - AB < BC

Hence, proved.

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