Question

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

Answer 1

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

To prove: AB + AC > AD

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

Proof:

Now in ADC and EDB,

AD = DE (by const)

DC = BD(as D is mid-point)

ADC = EDB (vertically opp. s)

Therefore,

In ABE, ADC EDB(by SAS)

This gives, BE = AC.

AB + BE > AE

AB + AC > 2AD ( AD = DE and BE = AC)

Hence the sum of any two sides of a triangle is greater than the median drawn to the third side

Answer 2

❌❌hello mates ❌❌

Given : Triangle ABC in which AD is the median.

To prove:AB+AC>2AD

Construction :

Extend AD to E such that AD=DE .

 Now join EC.

Proof:

In ΔADB and ΔEDC

AD=DE[ By construction]

D is the midpoint BC.[DB=DB]

ΔADB=ΔEDC [vertically opposite angles]

Therefore Δ ADB ≅  ΔEDC [ By SAS congruence

AB=ED[Corresponding parts of congruent triangles ]

In ΔAEC,

AC+ED> AE [sum of any two sides of a triangle is greater than the third side]

AC+AB>2AD[AE=AD+DE=AD+AD=2AD and ED=AB]