Question

SOLVE plz i have provided the question is attached in the picture​

Answer 1

Answer:

Extend the triangle ABC to a parallelogram such that AD=DK .

The triangles ADC and KDB are congruent (by SAS), so AC=BK .

Given that sum of the lengths of two sides of a triangles is greater than the third, we have

AK<AB+BK

AD+DK<AB+AC

2AD<AB+AC

Doing the same with the other medians,

we have,

2BE<BC+AB

2CF<AC+BC

Summing all three, we get,

2(AD+BE+CF)<2(AB+AC+BC)

or

AD+BE+CF<AB+AC+BC

Or in other words,

Sum of medians<Perimeter