Question

State whether the following statement is True or False.
If in a parallelogram ABCD, AC = BD and AC is perpendicular to BD, then ABCD is a square.

• A. True
• B. False

Answer 1

The correct option is A True

Given: In parallelogram $ABCD$, the diagonals $AC\&BD$ are equal & perpendicular to each other.

$\angle DOC=\angle COB=\angle BOA=\angle AOD={90}^o$

We know that diagonals of a parallelogram bisect each other.

Hence $OA=OC=OB=OD...\left(1\right)$

In $△AOD\&△AOB$,

$OA=OA$ [ Common side ]

$\angle AOD=\angle BOA={90}^o$ [ Given ]

$OD=OB$ [ from (1) ]

So $△AOD\cong △AOB$ [ By SAS Rule of Congruency ]

Therefore $AD=AB$.

So $AB=BC=CD=DA....\left(2\right)$ [ Since $ABCD$ is a parallelogram ]

Since all sides of the parallelogram are equal, it's a rhombous.

Since, diagonals of the rhombous $ABCD$ are equal, $ABCD$ is a square.