Respuesta :
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
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
![Ver imagen iuliaelisa15](https://us-static.z-dn.net/files/d92/9b5992121fa6a2a957e87ab7ef605e69.png)
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|
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|
![Ver imagen gmany](https://us-static.z-dn.net/files/da4/0dfa42dbeb83e4baac4cb2b986553afb.png)