In quadrilateral \(A B C D, \mathbf{A B}=\mathbf{a}, \mathbf{B C}=\mathbf{b}\), \(\mathbf{A…
In quadrilateral \(A B C D, \mathbf{A B}=\mathbf{a}, \mathbf{B C}=\mathbf{b}\), \(\mathbf{A D}=\mathbf{b}-\mathbf{a}\) if \(M\) is the midpoint of \(B C\) and \(N\) is a point on \(D M\) such that \(D N=\left(\frac{4}{5}\right) D M\), then \(5 \mathbf{A N}=\)
\(\mathrm{AC}\)
\(2 \mathrm{AC}\)
\(3 \mathrm{AC}\)
4AC
Solution
It is given that for a quadrilateral \(A B C D\),
\(\mathbf{A B}=\mathbf{a}, \mathbf{B C}=\mathbf{b}, \mathbf{A D}=\mathbf{b}-\mathbf{a}\)
\(\therefore\) According to the question, \(\mathbf{B M}=\frac{\mathbf{b}}{2}\) and
\(\begin{aligned}
& \frac{D N}{N M}=\frac{4}{1} \quad\left\{\because D N=\left(\frac{4}{5}\right) D M \text { given }\right\} \\
& \therefore \quad \mathbf{A N}-\mathbf{A D}=\frac{4}{1}(\mathbf{A M}-\mathbf{A N}) \\
& \Rightarrow \quad 5 \mathrm{AN}=4 \mathbf{A M}+\mathbf{A D}=4\left(\mathbf{a}+\frac{\mathbf{b}}{2}\right)+(\mathbf{b}-\mathbf{a}) \\
& =4 \mathbf{a}+2 \mathbf{b}+\mathbf{b}-\mathbf{a}=3(\mathbf{a}+\mathbf{b})=3 \mathbf{A C} \\
\end{aligned}\)
Hence, option (c) is correct.