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

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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|


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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


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