# Let us prove that Consider the parallelogram ABCD with diagonals AC and BD as shown and they intersect at right angles at O

A parallelogram is a square if its adjacent angles are 90° and adjacent sides are equal

So we have to prove that adjacent sides of given parallelogram are equal and adjacent angles are 90° to prove given parallelogram is a square

AC = BD … given

AB = DC … opposite sides of a parallelogram

Therefore, ΔBAD ΔCDA … SSS test for congruency

BAD = CDA …corresponding angles of congruent triangles …(i)

As it is given that ABCD is parallelogram

BAD + CDA = 180° … sum of adjacent angles of a parallelogram is 180°

Using equation (i)

CDA + CDA = 180°

2 × CDA = 180°

CDA = 90°

Thus, adjacent angles are 90° … (iii)

Consider ΔAOB and ΔAOD

AOB = AOD … both 90° because given that diagonals intersect at right angles

OD = OB … diagonals of a parallelogram bisect each other

AO is the common side

Therefore, ΔAOB ΔAOD … SAS test for congruency

AB = AD … corresponding sides of congruent triangles

Thus, adjacent sides are equal … (iv)

From statements (iii) and (iv) we can conclude that parallelogram ABCD is a square

Therefore, if in a parallelogram, the diagonals are equal in lengths and intersect at right angles, the parallelogram will be a square.

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