Question

prove that the difference between any two sides of a triangle is less than its third side​

Answer 1

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Answer 2

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$\huge{\underline{\underline{\sf{\orange{</p><p>Question:-}}}}}$

prove that the difference between any two sides of a triangle is less than its third side

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$\huge{\underline{\underline{\sf{\orange{</p><p>Solution:-}}}}}$

${\blue{\sf\underline{Given}}}$

A ∆ABC.

${\pink{\sf\underline{To\:Prove}}}$

(i)AC - AB < BC,

(ii)BC - AC < AB,

(iii)BC - AB < AC.

${\blue{\sf\underline{Construction}}}$

Let AC > AB. Then, along AC,set off AD = AB.Join BD.

${\pink{\sf\underline{Proof}}}$

AB = AD⇒ ∠1 = ∠2.⠀⠀⠀⠀⠀⠀⠀....(i)

Side CD of ∆BCD has been produced to A.

∴ ∠2 > ∠4⠀⠀⠀⠀...(ii) [∴ ext.angle > each int.opp angle]

Side AD of ∆ABD has been produced to C.

∴ ∠3 > ∠1⠀⠀⠀⠀....(iii) [ext.angle > each int.opp angle]

⇒∠3 > ∠2⠀⠀⠀⠀...(iv) [using (i)].

From (ii) and (iv),we get ∠3 > ∠4 ⇒ ∠4 < ∠3.

Now,∠4 < ∠3

⇒CD < BC

⇒AD - AD < BC

⇒AC-AB < BC⠀⠀[∴AD = AD].

Hence,AC-AB < BC.

Similarly,BC - AC < AB and BC - AB < AC.

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