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