Question

prove that the sum of any two sides of a triangle is greater than the third side​

Answer 1

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

prove that the sum of any two sides of a triangle is greater than the third side

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

A ∆ABC

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

(i)AB + AC > BC,

(ii)AB + BC > AC,

(iii)BC + AC > AB.

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

Produce BA to D such that AD = AC

Join CD.

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

(i) In ∆ACD,we have

AD = AC⠀⠀[by construction]

⇒∠ACD = ∠ADC⠀[opposite to equal sides]

⇒∠BCD > ∠ADC⠀[∴ ∠BCD > ∠ACD]

⇒∠BCD > ∠BDC⠀[∴ ∠ADC = ∠BDC]

⇒∠BD > ∠BC⠀⠀[side opposite to larger angle is larger]

⇒BA + AD > BC

⇒BA + AC > BC⠀[∴ AD = AC].

∴AB + AC > BC.

Similarly, AB + BC > AC and BC + AC > AB.

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Answer 2

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question - prove that the sum of any two sides of a triangle is greater than the third side

Solution -

Given - A ΔABC

To prove

(i)AB + AC > BC,

(i)AB + AC > BC,(ii)AB + BC > AC,

(i)AB + AC > BC,(ii)AB + BC > AC,(iii)BC + AC > AB.

Construction - Produce BA to D such that AD = AC

Join CD.

Prove ▬▬▬▬▬

(i) In ∆ACD,we have

(i) In ∆ACD,we haveAD = AC⠀⠀[by construction]

(i) In ∆ACD,we haveAD = AC⠀⠀[by construction]⇒∠ACD = ∠ADC⠀[opposite to equal sides]

(i) In ∆ACD,we haveAD = AC⠀⠀[by construction]⇒∠ACD = ∠ADC⠀[opposite to equal sides]⇒∠BCD > ∠ADC⠀[∴ ∠BCD > ∠ACD]

(i) In ∆ACD,we haveAD = AC⠀⠀[by construction]⇒∠ACD = ∠ADC⠀[opposite to equal sides]⇒∠BCD > ∠ADC⠀[∴ ∠BCD > ∠ACD]⇒∠BCD > ∠BDC⠀[∴ ∠ADC = ∠BDC]

(i) In ∆ACD,we haveAD = AC⠀⠀[by construction]⇒∠ACD = ∠ADC⠀[opposite to equal sides]⇒∠BCD > ∠ADC⠀[∴ ∠BCD > ∠ACD]⇒∠BCD > ∠BDC⠀[∴ ∠ADC = ∠BDC]⇒∠BD > ∠BC⠀⠀[side opposite to larger angle is larger]

(i) In ∆ACD,we haveAD = AC⠀⠀[by construction]⇒∠ACD = ∠ADC⠀[opposite to equal sides]⇒∠BCD > ∠ADC⠀[∴ ∠BCD > ∠ACD]⇒∠BCD > ∠BDC⠀[∴ ∠ADC = ∠BDC]⇒∠BD > ∠BC⠀⠀[side opposite to larger angle is larger]⇒BA + AD > BC

(i) In ∆ACD,we haveAD = AC⠀⠀[by construction]⇒∠ACD = ∠ADC⠀[opposite to equal sides]⇒∠BCD > ∠ADC⠀[∴ ∠BCD > ∠ACD]⇒∠BCD > ∠BDC⠀[∴ ∠ADC = ∠BDC]⇒∠BD > ∠BC⠀⠀[side opposite to larger angle is larger]⇒BA + AD > BC⇒BA + AC > BC⠀[∴ AD = AC].

(i) In ∆ACD,we haveAD = AC⠀⠀[by construction]⇒∠ACD = ∠ADC⠀[opposite to equal sides]⇒∠BCD > ∠ADC⠀[∴ ∠BCD > ∠ACD]⇒∠BCD > ∠BDC⠀[∴ ∠ADC = ∠BDC]⇒∠BD > ∠BC⠀⠀[side opposite to larger angle is larger]⇒BA + AD > BC⇒BA + AC > BC⠀[∴ AD = AC].∴AB + AC > BC.

(i) In ∆ACD,we haveAD = AC⠀⠀[by construction]⇒∠ACD = ∠ADC⠀[opposite to equal sides]⇒∠BCD > ∠ADC⠀[∴ ∠BCD > ∠ACD]⇒∠BCD > ∠BDC⠀[∴ ∠ADC = ∠BDC]⇒∠BD > ∠BC⠀⠀[side opposite to larger angle is larger]⇒BA + AD > BC⇒BA + AC > BC⠀[∴ AD = AC].∴AB + AC > BC.Similarly, AB + BC > AC and BC + AC > AB.

Hope it helps ❣️