Question

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

Answer 1

Answer:

Step-by-step explanation:

Let ABC be a triangle

We can extend BA past A into a straight line.

There exists a point D such that DA=CA.

Therefore, from Isosceles Triangle has Two Equal Angles:

∠ADC=∠ACD

Thus by Euclid's fifth common notion:

∠BCD>∠BDC

Since △DCB is a triangle having ∠BCD greater than ∠BDC, this means that BD>BC.

But:

BD=BA+AD

and:

AD=AC

Thus:

BA+AC>BC