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

Only authorized users can leave an answer! If you are not satisfied with the answer or you can’t find one, then try to use the search above or find similar answers below.

More questions               