Answer:
Proven
Step-by-step explanation:
Let ABC be a triangle and D, E and F are midpoints of BC, CA and AB.
The sum of two sides of a triangle is greater than twice the median bisecting the third side,
Hence in ΔABD, AD is a median
⇒ AB + AC > 2(AD)
Hence, we get
BC + AC > 2 (CF )
BC + AB > 2 (BE)
On adding the above inequalities, we get
(AB + AC) + (BC + AC) + (BC + AB )> 2 (AD) + 2 (CD) + 2 (BE )
2(AB + BC + AC) > 2(AD + BE + CF)
AB + BC + AC > AD + BE + CF - Proven
* Sorry I don't have any attachments, use your imagination!
