Question

In the quadrilateral $ABCD$, the diagonals $AC$ and $BD$ are equal and perpendicular to each other. What type of a quadrilateral is $ABCD$?

• A. A square
• B. A parallelogram
• C. A rectangle
• D. A trapezium

Answer 1

The correct option is A A square

$\Rightarrow$ In given figure ABCD is a square having all sides are equal and opposite sides parallel to each other. AC and BD are diagonals.

$\Rightarrow$ In $△$ABC and $△$BAD,

$\Rightarrow$ AB = AB [Common line]

$\Rightarrow$ BC = AD [Sides of square are equal.]

$\Rightarrow$ $\angle$ABC = $\angle$BAD [All four angles of square is 90${}^{\circ }$]

$\Rightarrow$ $△$ABC $\cong$ $△$BAD [By SAS property]

$\Rightarrow$ In a $△$ OAD and $△$OCB,

$\Rightarrow$ AD = CB [sides of a square]

$\Rightarrow$ $\angle$OAD = $\angle$OCB [Alternate angle]

$\Rightarrow$ $\angle$ODA = $\angle$OBC [Alternate angle]

$\Rightarrow$ $△$OAD $\cong$ $△$OCB [By ASA Property]

$\Rightarrow$ So, OA = OC ----- ( 1 )

$\Rightarrow$ Similarly, OB = OD ------ ( 2 )

From ( 1 ) and ( 2 ) we get that AC and BD bisect each other.

$\Rightarrow$ Now, in $△$OBA and $△$ODA,

$\Rightarrow$ OB = OD [From ( 2 )]

$\Rightarrow$ BA = DA

$\Rightarrow$ OA = OA [Common line]

$\Rightarrow$ $\angle$AOB + $\angle$AOD ----- ( 3 ) [By CPCT]

$\Rightarrow$ $\angle$AOB + $\angle$AOD = 180${}^{\circ }$ [Linear pair]

$\Rightarrow$ 2$\angle$AOB = 180${}^{\circ }$

$\therefore$ $\angle$AOB = $\angle$AOD = 90${}^{\circ }$

$\therefore$ We have proved that diagonals of square are equal and perpendicular to each other.