Question

In triangle ABC,AD is the median. Show that AB+BC+AC>2AD

Answer 1

To prove:AB+BC+AC>2AD Consider triangle ABD,AB+BD>AD.......(i) Consider triangle ADC, AC+CD>AD.......(ii) (i)+(ii)=› AB+AC+BD+DC>2AD [BD+DC=BC] So, AB+BC+BC>2AD Hence proved.

Answer 2

given,

ad is the median of Δabc

to prove,

ab + ac > 2ad

construction,

produce ad to E such that ad = de and join ce

ad = de(Construction)

∠adb = ∠cde(Vertically opposite angles)

bd = dc(ad is the median from A to BC)

∴ Δabd congruent to  Δcde by (SAS rule)  

⇒ ab = ce(cpct) ...(1)

In Δace

ac + ce > AE(Sum of any two sides of a triangle is greater than the third side)

⇒ ac + ab > ad + de [Using (1)]

⇒ ac + AB > ad + ad(Constriction)

⇒ ac + ab > 2ad {proved}