Question

Prove that any two sides of a triangle are together greater than twice the median drawn to the third side.​

Answer 1

Step-by-step explanation:

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Answer 2

⠀⠀⠀⠀⠀⠀⠀⠀⠀● Given: In ∆ABC, AD is the median To prove: AB + AC > 2AD

⠀⠀⠀⠀⠀⠀⠀⠀⠀● Construction: Produce AD up to E, such that AD + DE and Join EC.

⠀⠀⠀⠀⠀⠀⠀⠀⠀●Proof: In ∆ADB and ∆EDC AD = DE (by construction) BD = DC (given) and ∠ADB = ∠EDC (vertically opposite angles) ∆ADB = ∆EDC (by SAS congruency property) ⇒ AB = CE (by c.p.c.t)

⠀⠀⠀⠀⠀⠀⠀⠀⠀● Now in ∆ACE AC + CE > AE ⇒ AC + AB > AE (∵ AB = CE) (Proved above) ⇒ AC + AB > 2AD (∵ AE = 2AD)

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