Question

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Answer 1

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

So, we have to prove that CD<BC

Proof:

Consider traingles ABD,

Since AB=AD, since opposite to the equal opposite sides are always equal. So, =P=P

Now let , using exterior angle property in traingle ABD, Exterior angle property of a triangle states that the measures of an exterior angles is equal to the sum of the two interior angles

(equation 1)

Now , using exterior angle property in triangle BCD,

(equation2)

Now, by comparing 1 and 2 So, x<y

CD<BC

CD<BC

AC- AD < BC

since , AD =AB

Therefore, AC -AB <BC

Hence , proved.

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