ABCDE is a pentagon, prove that vector(AB) + vector(AE) + vector(BC) + vector(DC) + vector(ED) + vector(AC) = 3vector(AC)

Question

ABCDE is a pentagon, prove that

\(\overset{\longrightarrow} {AB}+\overset{\longrightarrow}{AE}+\overset{\longrightarrow} {BC}+\overset{\longrightarrow}{DC}+\overset{\longrightarrow} {ED}+\overset{\longrightarrow} {AC}=3\overset{\longrightarrow} {AC} \)

Answer 1

Given: ABCDE is a pentagon as shown below

Consider ΔABC and apply triangle law of vector, we get

\(\overset{\longrightarrow}{AB}+\overset{\longrightarrow}{BC}=\overset{\longrightarrow}{AC} \)....(i)

Similarly, consider ΔADE and apply triangle law of vector, we get

\(\overset{\longrightarrow}{AE}+\overset{\longrightarrow}{ED}=\overset{\longrightarrow}{AD} \)....(ii)

And, consider ΔADC and apply triangle law of vector, we get

\(\overset{\longrightarrow}{AD}+\overset{\longrightarrow}{DC}=\overset{\longrightarrow}{AC} \)....(iii)

Adding (i), (ii) and (iii), we get

Hence proved.