Question

prove that in a triangle the sum of the medians is less than the perimeter​

Answer 1

Step-by-step explanation:

Let ABC be the triangle and D. E and F are midpoints of BC, CA and AB respectively.

Recall that 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)

Similarly, we get

BC + AC > 2CF

BC + AB > 2BE

On adding the above inequations, we get

(AB + AC) + (BC + AC) + (BC + AB )> 2AD + 2CD + 2BE

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

∴ AB + BC + AC > AD + BE + CF

Answer 2

Answer:

Brain list answer

Step-by-step explanation:

In ∆PQS

(PQ+ QS) is greater than PS -------- EQ -1

In ∆PSR

(PR +RS) is greater than PS -------EQ-2

adding eq-1and eq-2

(PQ+QS+PR+RS) is greater than 2PS --EQ-3

LLY, WE GET

( PQ+QS+PR) is greater than 2RW ---EQ-4

(PR+PQ+QR) is greater than 2QT ---EQ-5

Adding 3,4and5;

3(PR+PQ+QR) is greater than 2(PS+RW+QT) = 3/2 (PR+PQ+QR) is greater than (PS+RW+QT)

PROVED