Question

If \[\vec{a}\], \[\vec{b}\], \[\vec{c}\] and \[\vec{d}\] are the position vectors of points A, B, C, D such that no three of them are collinear and \[\vec{a} + \vec{c} = \vec{b} + \vec{d} ,\] then ABCD is a

विकल्प

•  rhombus

• rectangle

• square

• parallelogram

Answer 1

Given:

\[\vec{a} + \vec{c} = \vec{b} + \vec{d} \]

\[ \Rightarrow \vec{c} - \vec{d} = \vec{b} - \vec{a} \]

\[ \Rightarrow \overrightarrow{AB} = \overrightarrow{DC} \]

\[\text{ And }\vec{a} + \vec{c} = \vec{b} + \vec{d} \]

\[ \Rightarrow \vec{c} - \vec{b} = \vec{d} - \vec{a} \]

\[ \Rightarrow \overrightarrow{AD} = \overrightarrow{BC} \]

\[\text{ Also, since }\vec{a} + \vec{c} = \vec{b} + \vec{d} \]

\[ \Rightarrow \frac{1}{2}( \vec{a} + \vec{c)} = \frac{1}{2}( \vec{b} + \vec{d} )\]

so, position vector of mid point of BD = position vector of mid point of AC .]

hence diagonals bisect each other .

the given ABCD is a paralle log ram .