A straight line $L$ cuts the lines $AB,ACandAD$ of a parallelogram $ABCD$ at points $B_1,C_{1}and{D}_1,$ respectively. If ${\left(\vec{A}B\right)}_1,{\lambda }_1\vec{A}B,{\left(\vec{A}D\right)}_1={\lambda }_2\vec{A}Dand{\left(\vec{A}C\right)}_1={\lambda }_3\vec{A}C,$ then prove that $\frac{1}{{\lambda }_3}=\frac{1}{{\lambda }_1}+\frac{1}{{\lambda }_2}$ .

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