\displaystyle \textbf{Question 1: }~\text{In a triangle }OAB,\ \angle AOB=90^\circ.\ \text{If }P\text{ and }Q \\ \text{ are points of trisection of }AB,\ \text{prove that }OP^{2}+OQ^{2}=\frac{5}{9}AB^{2}.
\displaystyle \text{Answer:}

\displaystyle \text{In }\triangle OAB,\ \angle AOB=90^\circ.\ P\text{ and }Q\text{ are points of trisection of }AB
\displaystyle \text{Taking }O\text{ as the origin, let the position vectors of }A\text{ and }B\text{ be }\overrightarrow{a}\text{ and }\overrightarrow{b}\text{ respectively}
\displaystyle \text{Since }P\text{ and }Q\text{ are the points of trisection of }AB,\ \text{so }AP:PB=1:2\text{ and }AQ:QB=2:1
\displaystyle \text{Position vector of }P,\ \overrightarrow{OP}=\frac{2\overrightarrow{a}+\overrightarrow{b}}{3}\ \text{(Using section formula)}
\displaystyle \text{Position vector of }Q,\ \overrightarrow{OQ}=\frac{\overrightarrow{a}+2\overrightarrow{b}}{3}
\displaystyle \therefore \overrightarrow{a}\cdot\overrightarrow{b}=0\ \text{...(1)}
\displaystyle \text{Now,}
\displaystyle OP^{2}+OQ^{2}
\displaystyle =|\overrightarrow{OP}|^{2}+|\overrightarrow{OQ}|^{2}
\displaystyle =\left(\frac{2\overrightarrow{a}+\overrightarrow{b}}{3}\right)\cdot\left(\frac{2\overrightarrow{a}+\overrightarrow{b}}{3}\right)+\left(\frac{\overrightarrow{a}+2\overrightarrow{b}}{3}\right)\cdot\left(\frac{\overrightarrow{a}+2\overrightarrow{b}}{3}\right)
\displaystyle =\frac{4|\overrightarrow{a}|^{2}+4\overrightarrow{a}\cdot\overrightarrow{b}+|\overrightarrow{b}|^{2}+|\overrightarrow{a}|^{2}+4\overrightarrow{a}\cdot\overrightarrow{b}+4|\overrightarrow{b}|^{2}}{9}
\displaystyle =\frac{5|\overrightarrow{a}|^{2}+5|\overrightarrow{b}|^{2}}{9}\ \text{[Using (1)]}
\displaystyle =\frac{5}{9}\left(|\overrightarrow{a}|^{2}+|\overrightarrow{b}|^{2}\right)
\displaystyle =\frac{5}{9}|\overrightarrow{AB}|^{2}\ \text{[Using Pythagoras Theorem]}
\displaystyle =\frac{5}{9}AB^{2}

\displaystyle \textbf{Question 2: }~\text{Prove that if the diagonals of a quadrilateral bisect each other at} \\ \text{right angles, then it is a rhombus.}
\displaystyle \text{Answer:}
\displaystyle \text{Taking }O\text{ as the origin, let the position vectors of }A\text{ and }B\text{ be }\overrightarrow{a}\text{ and }\overrightarrow{b}\text{ respectively.}
\displaystyle \text{Then, }\overrightarrow{OA}=\overrightarrow{a}\text{ and }\overrightarrow{OB}=\overrightarrow{b}
\displaystyle \text{Position vector of the mid-point of }AB,\ \overrightarrow{OE}=\frac{\overrightarrow{a}+\overrightarrow{b}}{2}
\displaystyle \therefore \text{Position vector of }C,\ \overrightarrow{OC}=\overrightarrow{a}+\overrightarrow{b}
\displaystyle \text{By the triangle law of vector addition, we have}
\displaystyle \overrightarrow{OA}+\overrightarrow{AB}=\overrightarrow{OB}
\displaystyle \Rightarrow \overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}=\overrightarrow{b}-\overrightarrow{a}
\displaystyle \text{Since }\overrightarrow{AB}\perp\overrightarrow{OC}
\displaystyle \Rightarrow \overrightarrow{AB}\cdot\overrightarrow{OC}=0
\displaystyle \Rightarrow (\overrightarrow{b}-\overrightarrow{a})\cdot(\overrightarrow{a}+\overrightarrow{b})=0
\displaystyle \Rightarrow |\overrightarrow{b}|^{2}-|\overrightarrow{a}|^{2}=0
\displaystyle \Rightarrow |\overrightarrow{a}|^{2}=|\overrightarrow{b}|^{2}
\displaystyle \Rightarrow |\overrightarrow{a}|=|\overrightarrow{b}|
\displaystyle \Rightarrow OA=OB
\displaystyle \text{In a quadrilateral, if the diagonals bisect each other at right angles and adjacent} \\ \text{sides are equal, then it is a rhombus.}

\displaystyle \textbf{Question 3: }~\text{(Pythagoras' Theorem) Prove by vector method that in a right angled triangle,} \\ \text{the square of the hypotenuse is equal to the sum of the squares of the other two sides.}
\displaystyle \text{Answer:}

\displaystyle \text{Let }ABC\text{ be a right triangle with }\angle BAC = 90^\circ.
\displaystyle \text{Taking }A\text{ as the origin, let the position vectors of }B\text{ and }C\text{ be }\overrightarrow{b}\text{ and }\overrightarrow{c}\text{ respectively.}
\displaystyle \text{Then, }\overrightarrow{AB}=\overrightarrow{b}\text{ and }\overrightarrow{AC}=\overrightarrow{c}.\displaystyle \text{Since }\overrightarrow{AB}\perp\overrightarrow{AC}
\displaystyle \Rightarrow \overrightarrow{b}\cdot\overrightarrow{c}=0 \qquad \ldots(1)\displaystyle \text{Now,}
\displaystyle |\overrightarrow{AB}|^{2}+|\overrightarrow{AC}|^{2}=|\overrightarrow{b}|^{2}+|\overrightarrow{c}|^{2} \qquad \ldots(2)\displaystyle \text{Also,}
\displaystyle |\overrightarrow{BC}|^{2}=|\overrightarrow{c}-\overrightarrow{b}|^{2}
\displaystyle =(\overrightarrow{c}-\overrightarrow{b})\cdot(\overrightarrow{c}-\overrightarrow{b})
\displaystyle =|\overrightarrow{c}|^{2}-2\,\overrightarrow{b}\cdot\overrightarrow{c}+|\overrightarrow{b}|^{2}
\displaystyle =|\overrightarrow{c}|^{2}+|\overrightarrow{b}|^{2} \qquad \ldots(3)\ \text{[Using (1)]}\displaystyle \text{From (2) and (3), we have}
\displaystyle |\overrightarrow{AB}|^{2}+|\overrightarrow{AC}|^{2}=|\overrightarrow{BC}|^{2}

\displaystyle \textbf{Question 4: }~\text{Prove by vector method that the sum of the squares of the diagonals of a} \\ \text{parallelogram is equal to the sum of the squares of its sides.}
\displaystyle \text{Answer:}

\displaystyle \text{Let }ABCD\text{ be a parallelogram such that }AC\text{ and }BD\text{ are its diagonals.}
\displaystyle \text{Taking }A\text{ as the origin, let the position vectors of }B\text{ and }D\text{ be }\overrightarrow{b}\text{ and }\overrightarrow{d}\text{ respectively.}

\displaystyle \text{Then, }\overrightarrow{AB}=\overrightarrow{b}\text{ and }\overrightarrow{AD}=\overrightarrow{d}.

\displaystyle \text{Using triangle law of vector addition, we have}
\displaystyle \overrightarrow{AD}+\overrightarrow{DB}=\overrightarrow{AB}
\displaystyle \Rightarrow \overrightarrow{DB}=\overrightarrow{b}-\overrightarrow{d}.

\displaystyle \text{In }\triangle ABC,
\displaystyle \overrightarrow{AC}=\overrightarrow{AB}+\overrightarrow{BC}=\overrightarrow{AB}+\overrightarrow{AD}=\overrightarrow{b}+\overrightarrow{d}.

\displaystyle \text{Now,}
\displaystyle |\overrightarrow{AB}|^{2}+|\overrightarrow{BC}|^{2}+|\overrightarrow{CD}|^{2}+|\overrightarrow{DA}|^{2}
\displaystyle =|\overrightarrow{AB}|^{2}+|\overrightarrow{AD}|^{2}+|\!-\!\overrightarrow{AB}|^{2}+|\!-\!\overrightarrow{AD}|^{2}
\displaystyle =2|\overrightarrow{AB}|^{2}+2|\overrightarrow{AD}|^{2}
\displaystyle =2|\overrightarrow{b}|^{2}+2|\overrightarrow{d}|^{2}\qquad \ldots(1)

\displaystyle \text{Also,}
\displaystyle |\overrightarrow{DB}|^{2}+|\overrightarrow{AC}|^{2}
\displaystyle =|\overrightarrow{b}-\overrightarrow{d}|^{2}+|\overrightarrow{b}+\overrightarrow{d}|^{2}
\displaystyle =(\overrightarrow{b}-\overrightarrow{d})\cdot(\overrightarrow{b}-\overrightarrow{d})+(\overrightarrow{b}+\overrightarrow{d})\cdot(\overrightarrow{b}+\overrightarrow{d})
\displaystyle =|\overrightarrow{b}|^{2}-2\overrightarrow{b}\cdot\overrightarrow{d}+|\overrightarrow{d}|^{2}+|\overrightarrow{b}|^{2}+2\overrightarrow{b}\cdot\overrightarrow{d}+|\overrightarrow{d}|^{2}
\displaystyle =2|\overrightarrow{b}|^{2}+2|\overrightarrow{d}|^{2}\qquad \ldots(2)

\displaystyle \text{From (1) and (2), we have}
\displaystyle |\overrightarrow{AB}|^{2}+|\overrightarrow{BC}|^{2}+|\overrightarrow{CD}|^{2}+|\overrightarrow{DA}|^{2} =|\overrightarrow{DB}|^{2}+|\overrightarrow{AC}|^{2}

\displaystyle \textbf{Question 5: }~\text{Prove using vectors that the quadrilateral obtained by joining the} \\ \text{mid-points of adjacent sides of a rectangle is a rhombus.}
\displaystyle \text{Answer:}

\displaystyle \text{ABCD is a rectangle. Let }P,Q,R\text{ and }S\text{ be the mid-points of sides } \\ AB,BC,CD\text{ and }DA\text{ respectively.}

\displaystyle \text{Now,}

\displaystyle \overrightarrow{PQ}=\overrightarrow{PB}+\overrightarrow{BQ} =\frac12\overrightarrow{AB}+\frac12\overrightarrow{BC} =\frac12(\overrightarrow{AB}+\overrightarrow{BC}) =\frac12\overrightarrow{AC}\qquad\ldots(1)

\displaystyle \overrightarrow{SR}=\overrightarrow{SD}+\overrightarrow{DR} =\frac12\overrightarrow{AD}+\frac12\overrightarrow{DC} =\frac12(\overrightarrow{AD}+\overrightarrow{DC}) =\frac12\overrightarrow{AC}\qquad\ldots(2)

\displaystyle \text{From (1) and (2), we have}
\displaystyle \overrightarrow{PQ}=\overrightarrow{SR}.

\displaystyle \text{So, the sides }PQ\text{ and }SR\text{ are equal and parallel. Thus, }PQRS\text{ is a parallelogram.}

\displaystyle \text{Now,}

\displaystyle |\overrightarrow{PQ}|^{2} =\overrightarrow{PQ}\cdot\overrightarrow{PQ}

\displaystyle \Rightarrow |\overrightarrow{PQ}|^{2} =(\overrightarrow{PB}+\overrightarrow{BQ})\cdot(\overrightarrow{PB}+\overrightarrow{BQ})

\displaystyle \Rightarrow |\overrightarrow{PQ}|^{2} =|\overrightarrow{PB}|^{2}+2\overrightarrow{PB}\cdot\overrightarrow{BQ}+|\overrightarrow{BQ}|^{2}

\displaystyle \Rightarrow |\overrightarrow{PQ}|^{2} =|\overrightarrow{PB}|^{2}+0+|\overrightarrow{BQ}|^{2}\qquad(\overrightarrow{PB}\perp\overrightarrow{BQ})

\displaystyle \Rightarrow |\overrightarrow{PQ}|^{2} =|\overrightarrow{PB}|^{2}+|\overrightarrow{BQ}|^{2}\qquad\ldots(3)

\displaystyle \text{Also,}

\displaystyle |\overrightarrow{PS}|^{2} =\overrightarrow{PS}\cdot\overrightarrow{PS}

\displaystyle \Rightarrow |\overrightarrow{PS}|^{2} =(\overrightarrow{PA}+\overrightarrow{AS})\cdot(\overrightarrow{PA}+\overrightarrow{AS})

\displaystyle \Rightarrow |\overrightarrow{PS}|^{2} =|\overrightarrow{PA}|^{2}+2\overrightarrow{PA}\cdot\overrightarrow{AS}+|\overrightarrow{AS}|^{2}

\displaystyle \Rightarrow |\overrightarrow{PS}|^{2} =|\overrightarrow{PB}|^{2}+0+|\overrightarrow{BQ}|^{2}\qquad(\overrightarrow{PA}\perp\overrightarrow{AS})

\displaystyle \Rightarrow |\overrightarrow{PS}|^{2} =|\overrightarrow{PB}|^{2}+|\overrightarrow{BQ}|^{2}\qquad\ldots(4)

\displaystyle \text{From (3) and (4), we have}
\displaystyle |\overrightarrow{PQ}|^{2}=|\overrightarrow{PS}|^{2}

\displaystyle \Rightarrow |\overrightarrow{PQ}|=|\overrightarrow{PS}|.

\displaystyle \text{So, the adjacent sides of the parallelogram are equal. Hence, }PQRS\text{ is a rhombus.}

\displaystyle \textbf{Question 6: }~\text{Prove that the diagonals of a rhombus are perpendicular bisectors} \\ \text{of each other.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }OABC\text{ be a rhombus, whose diagonals }OB\text{ and }AC\text{ intersect at }D.\ \text{Suppose }O\text{ is the origin.}
\displaystyle \text{Let the position vectors of }A\text{ and }C\text{ be }\overrightarrow{a}\text{ and }\overrightarrow{c}\text{ respectively.}
\displaystyle \text{Then, }\overrightarrow{OA}=\overrightarrow{a}
\displaystyle \text{In }\triangle OAB,
\displaystyle \overrightarrow{OB}=\overrightarrow{OA}+\overrightarrow{AB}=\overrightarrow{OA}+\overrightarrow{OC}=\overrightarrow{a}+\overrightarrow{c}\qquad(\overrightarrow{AB}=\overrightarrow{OC})
\displaystyle \text{Position vector of mid-point of }\overrightarrow{OB}=\frac{1}{2}(\overrightarrow{a}+\overrightarrow{c})
\displaystyle \text{Position vector of mid-point of }\overrightarrow{AC}=\frac{1}{2}(\overrightarrow{a}+\overrightarrow{c})\qquad\text{(Mid-point formula)}
\displaystyle \text{So, the mid-points of }OB\text{ and }AC\text{ coincide. Thus, the diagonals }OB\text{ and }AC\text{ bisect each other.}
\displaystyle \text{Now,}
\displaystyle \overrightarrow{OB}\cdot\overrightarrow{AC}=(\overrightarrow{a}+\overrightarrow{c})\cdot(\overrightarrow{c}-\overrightarrow{a})
\displaystyle =(\overrightarrow{c}+\overrightarrow{a})\cdot(\overrightarrow{c}-\overrightarrow{a})
\displaystyle =|\overrightarrow{c}|^{2}-|\overrightarrow{a}|^{2}
\displaystyle =|\overrightarrow{OC}|^{2}-|\overrightarrow{OA}|^{2}
\displaystyle =0\qquad(|\overrightarrow{OC}|=|\overrightarrow{OA}|)
\displaystyle \Rightarrow \overrightarrow{OB}\perp\overrightarrow{AC}
\displaystyle \text{Hence, the diagonals }OB\text{ and }AC\text{ are perpendicular to each other.}

\displaystyle \textbf{Question 7: }~\text{Prove that the diagonals of a rectangle are perpendicular if and} \\ \text{only if the rectangle is a square.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }ABCD\text{ be a rectangle. Take }A\text{ as the origin.}
\displaystyle \text{Suppose the position vectors of points }B\text{ and }D\text{ be }\overrightarrow{a}\text{ and }\overrightarrow{b}\text{ respectively.}
\displaystyle \text{Now,}
\displaystyle \overrightarrow{AC}=\overrightarrow{AB}+\overrightarrow{BC}=\overrightarrow{AB}+\overrightarrow{AD}=\overrightarrow{a}+\overrightarrow{b}
\displaystyle \text{Also,}
\displaystyle \overrightarrow{BD}=\overrightarrow{D}-\overrightarrow{B}=\overrightarrow{b}-\overrightarrow{a}
\displaystyle \text{Since }ABCD\text{ is a rectangle, }\overrightarrow{AB}\perp\overrightarrow{AD}
\displaystyle \Rightarrow \overrightarrow{a}\cdot\overrightarrow{b}=0
\displaystyle \text{Now, the diagonals }AC\text{ and }BD\text{ are perpendicular if and only if } \overrightarrow{AC}\cdot\overrightarrow{BD}=0
\displaystyle \iff (\overrightarrow{a}+\overrightarrow{b})\cdot(\overrightarrow{b}-\overrightarrow{a})=0
\displaystyle \iff |\overrightarrow{b}|^{2}-|\overrightarrow{a}|^{2}=0
\displaystyle \iff |\overrightarrow{a}|=|\overrightarrow{b}|
\displaystyle \iff |\overrightarrow{AB}|=|\overrightarrow{AD}|
\displaystyle \iff ABCD\text{ is a square.}
\displaystyle \text{Thus, the diagonals of a rectangle are perpendicular if and only if the rectangle is a square.}

\displaystyle \textbf{Question 8: }~\text{If }AD\text{ is the median of }\triangle ABC,\ \text{using vectors, prove that } \\ AB^{2}+AC^{2}=2(AD^{2}+CD^{2}).
\displaystyle \text{Answer:}
\displaystyle \text{Taking }A\text{ as the origin, let the position vectors of }B\text{ and }C\text{ be }\overrightarrow{b}\text{ and }\overrightarrow{c}\text{ respectively.}
\displaystyle \text{It is given that }AD\text{ is the median of }\triangle ABC.
\displaystyle \therefore \text{Position vector of the mid-point of }BC=\overrightarrow{AD}=\frac{\overrightarrow{b}+\overrightarrow{c}}{2}\qquad\text{(Mid-point formula)}
\displaystyle \text{Now,}


\displaystyle AB^{2}+AC^{2}=|\overrightarrow{AB}|^{2}+|\overrightarrow{AC}|^{2}=|\overrightarrow{b}|^{2}+|\overrightarrow{c}|^{2}\qquad\ldots(1)
\displaystyle \text{Also,}
\displaystyle 2(AD^{2}+CD^{2})=2\left(|\overrightarrow{AD}|^{2}+|\overrightarrow{CD}|^{2}\right)
\displaystyle =2\left[\left(\frac{\overrightarrow{b}+\overrightarrow{c}}{2}\right)\cdot\left(\frac{\overrightarrow{b}+\overrightarrow{c}}{2}\right) +\left(\frac{\overrightarrow{b}+\overrightarrow{c}}{2}-\overrightarrow{c}\right)\cdot\left(\frac{\overrightarrow{b}+\overrightarrow{c}}{2}-\overrightarrow{c}\right)\right]
\displaystyle =2\left[\left(\frac{\overrightarrow{b}+\overrightarrow{c}}{2}\right)\cdot\left(\frac{\overrightarrow{b}+\overrightarrow{c}}{2}\right) +\left(\frac{\overrightarrow{b}-\overrightarrow{c}}{2}\right)\cdot\left(\frac{\overrightarrow{b}-\overrightarrow{c}}{2}\right)\right]
\displaystyle =\frac{|\overrightarrow{b}|^{2}+2\overrightarrow{b}\cdot\overrightarrow{c}+|\overrightarrow{c}|^{2}}{2} +\frac{|\overrightarrow{b}|^{2}-2\overrightarrow{b}\cdot\overrightarrow{c}+|\overrightarrow{c}|^{2}}{2}
\displaystyle =\frac{2|\overrightarrow{b}|^{2}+2|\overrightarrow{c}|^{2}}{2}
\displaystyle =|\overrightarrow{b}|^{2}+|\overrightarrow{c}|^{2}\qquad\ldots(2)
\displaystyle \text{From (1) and (2), we have}
\displaystyle AB^{2}+AC^{2}=2(AD^{2}+CD^{2})

\displaystyle \textbf{Question 9: }~\text{If the median to the base of a triangle is perpendicular to the base,} \\ \text{then prove that the triangle is isosceles.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }\triangle ABC\text{ be a triangle such that }AD\text{ is a median and }AD\perp BC.\text{ Taking }A \\ \text{ as the origin, let the position vectors of }B\text{ and }C\text{ be }\overrightarrow{b}\text{ and }\overrightarrow{c}\text{ respectively.}
\displaystyle \text{Then the position vector of }D=\frac{\overrightarrow{b}+\overrightarrow{c}}{2}\text{ (mid-point formula).}
\displaystyle \overrightarrow{AD}=\text{position vector of }D-\text{position vector of }A=\frac{\overrightarrow{b}+\overrightarrow{c}}{2}
\displaystyle \overrightarrow{BC}=\text{position vector of }C-\text{position vector of }B=\overrightarrow{c}-\overrightarrow{b}
\displaystyle \text{Since }\overrightarrow{AD}\perp\overrightarrow{BC},\ \overrightarrow{AD}\cdot\overrightarrow{BC}=0
\displaystyle \frac{1}{2}(\overrightarrow{b}+\overrightarrow{c})\cdot(\overrightarrow{c}-\overrightarrow{b})=0
\displaystyle (\overrightarrow{b}+\overrightarrow{c})\cdot(\overrightarrow{c}-\overrightarrow{b})=0
\displaystyle \overrightarrow{c}\cdot\overrightarrow{c}-\overrightarrow{b}\cdot\overrightarrow{b}=0
\displaystyle |\overrightarrow{c}|^{2}-|\overrightarrow{b}|^{2}=0
\displaystyle |\overrightarrow{c}|=|\overrightarrow{b}|
\displaystyle AC=AB
\displaystyle \text{Hence, }\triangle ABC\text{ is an isosceles triangle.}

\displaystyle \textbf{Question 10: }~\text{In a quadrilateral }ABCD,\ \text{prove that }AB^{2}+BC^{2}+CD^{2}+DA^{2}=AC^{2}+BD^{2}+4PQ^{2}, \text{where }P\text{ and }Q \text{ are the mid-points of diagonals } \\ AC\text{ and }BD\text{ respectively.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }ABCD\text{ be a quadrilateral. Taking }A\text{ as the origin, let the position vectors of }B,C\text{ and }D \\ \text{ be }\overrightarrow{b},\overrightarrow{c}\text{ and }\overrightarrow{d}\text{ respectively.}
\displaystyle \text{Then the position vector of }P=\frac{\overrightarrow{c}}{2}\text{ (mid-point formula).}
\displaystyle \text{The position vector of }Q=\frac{\overrightarrow{b}+\overrightarrow{d}}{2}\text{ (mid-point formula).}

\displaystyle AB^{2}+BC^{2}+CD^{2}+DA^{2}
\displaystyle =|\overrightarrow{AB}|^{2}+|\overrightarrow{BC}|^{2}+|\overrightarrow{CD}|^{2}+|\overrightarrow{DA}|^{2}
\displaystyle =|\overrightarrow{b}|^{2}+|\overrightarrow{c}-\overrightarrow{b}|^{2}+|\overrightarrow{d}-\overrightarrow{c}|^{2}+|\overrightarrow{d}|^{2}
\displaystyle =|\overrightarrow{b}|^{2}+|\overrightarrow{c}|^{2}-2\overrightarrow{c}\cdot\overrightarrow{b}+|\overrightarrow{b}|^{2}+|\overrightarrow{d}|^{2}-2\overrightarrow{d}\cdot\overrightarrow{c}+|\overrightarrow{c}|^{2}+|\overrightarrow{d}|^{2}
\displaystyle =2|\overrightarrow{b}|^{2}+2|\overrightarrow{c}|^{2}+2|\overrightarrow{d}|^{2}-2\overrightarrow{b}\cdot\overrightarrow{c}-2\overrightarrow{c}\cdot\overrightarrow{d}\qquad (1)
\displaystyle AC^{2}+BD^{2}+4PQ^{2}
\displaystyle =|\overrightarrow{AC}|^{2}+|\overrightarrow{BD}|^{2}+4|\overrightarrow{PQ}|^{2}
\displaystyle =|\overrightarrow{c}|^{2}+|\overrightarrow{d}-\overrightarrow{b}|^{2}+4\left|\frac{\overrightarrow{b}+\overrightarrow{d}-\overrightarrow{c}}{2}\right|^{2}
\displaystyle =|\overrightarrow{c}|^{2}+|\overrightarrow{d}|^{2}+|\overrightarrow{b}|^{2}-2\overrightarrow{b}\cdot\overrightarrow{d}+|\overrightarrow{b}+\overrightarrow{d}|^{2}-2(\overrightarrow{b}+\overrightarrow{d})\cdot\overrightarrow{c}+|\overrightarrow{c}|^{2}
\displaystyle =2|\overrightarrow{b}|^{2}+2|\overrightarrow{c}|^{2}+2|\overrightarrow{d}|^{2}-2\overrightarrow{b}\cdot\overrightarrow{c}-2\overrightarrow{c}\cdot\overrightarrow{d}\qquad (2)
\displaystyle \text{From (1) and (2), we have }AB^{2}+BC^{2}+CD^{2}+DA^{2}=AC^{2}+BD^{2}+4PQ^{2}

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