Question

prove that the sum of any two sides of a triangle is greater than twice the median drawn to the third side.​

Answer 1

Step-by-step explanation:

Given: In △ABC,AD is the median drawn from A to BC

To Prove that AB+AC>AD

Construction:Produce AD to E so that DE=AD, join BE

Proof:In △ADC and △EDB we have

AD=DE(constant)

DC=BD as D is the midpoint

∠ADC=∠EDB (vertically opposite angles)

∴ In △ABE, △ADC≅△EDB by S.A.S

This gives BE=AC

AB+BE>AE

AB+AC>2AD ∵AD=DE and BE=AC

Hence the sum of any two sides

• mark as brilliant
• please draw a diagram according to my solution

Answer 2

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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 twice the median drawn to the third side.

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

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

A ∆ABC in which AD is a median.

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

AB + AC > 2AD.

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

Produce AD to E such that AD = DE.Join EC.

${\green{\sf\underline{Proof}}}$

In ∆ADB and ∆EDC,we have

AD = DE⠀(by construction)

∠ADB = ∠EDC⠀(vert.opp. angles)

BD = CD⠀(∴ D is the midpoint of BC)

∴ ∆ADB ≅ ∆EDC⠀(by SAS-criteria)

∴ AB = EC⠀(c.p.c.t).

We know that the sum of any two sides of a triangle is greater than the third side.So, in ∆ACE,we have

EC + AC > AE.

∴AB + AC > 2AD⠀[∴EC = AB and AE = 2AD].

Hence,AB + AC > 2AD.

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