ABCDE is a pentagon. prove that AB+AE+BC+DC+ED=2AC

Hi

 Triangle law of vector addition

If two vectors are represented by two sides of a triangle in sequence, then third closing side of the triangle, in the opposite direction of the sequence, represents the sum (or resultant) of the two vectors in both magnitude and direction.

Here, the term “sequence” means that the vectors are placed such that tail of a vector begins at the arrow head of the vector placed before it.

Figure 3:

Triangle law of vector addition

The triangle law does not restrict where to start i.e. with which vector to start. Also, it does not put conditions with regard to any specific direction for the sequence of vectors, like clockwise or anti-clockwise, to be maintained. In figure (i), the law is applied starting with vector,b; whereas the law is applied starting with vector, a, in figure (ii). In either case, the resultant vector, c, is same in magnitude and direction.

Now divide the given pentagon in three parts by joining AB and AD. Consider triangle ABC , by triangle law, AB + BC = AC (1)

for triangle ACD , AC + CD = AD (2)

For triangle ADE , AD + DE = AE  (3)

From 2 and 3 we get  AC + CD = AE - DE .=> AC= AE + ED + DC (4)

Add (1) and (4) to get the desired answer.