Question

Points A,B, and C are collinear. Points M and N are the midpoints of segments AB and AC. Prove that BC = 2MN

Answer 1

Look at the picture.

1)|AM| = |MB| = x
|AN| = |NC| = y
|BC| = 2y - 2x = 2(y - x)
|MN| = y - x
Therefore |BC| = 2|MN|
2)|AM| = |MB| = x
|AN| = |NC| = y

|BC| = 2y - 2x = 2(y - x)
|MN| = y - x
Therefore |BC| = 2|MN|

Answer 2

Hello ! (I attached a picture of the drawing)

First, we draw a line with 3 points B,A,C in this order.Then we will note the middle of AB with M and also the middle of AC with N.

We can notice that BC = AB + AC

M - the middle of AB ⇒ AM=MB=AB/2
N - the middle of AC ⇒ AN=NC=AC/2


We can write MN as AM+AN , which means AB/2 + AC/2 . We can also write AB as BC-AC , so we will have :

MN = AB/2 + AC/2
MN = (BC-AC)/2 + AC/2
MN = BC/2 - AC/2 + AC/2 ( AC/2 with -AC/2 are reduced)
MN = BC/2    or    BC = 2·MN