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# Show that if the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square.

Given: A quadrilateral ABCD in which AC = BD and AC $$\perp{BD}$$ such that OA = OC and OB = OD. So, ABCD is a parallelogram.

To prove: ABCD is a square.

Proof: Let AC and BD intersect at a point O.
In $$\triangle{ABO}$$ and $$\triangle{ADO}$$, we have,
BO = OD ...(given)
AO = OA ...(Common side)
$$\angle{AOB}$$ = $$\angle{AOD}$$ = $$90^\circ$$ ...(given)
$$\therefore$$ $$\triangle{ABO}$$ $$\displaystyle \cong$$ $$\triangle{ADO}$$
(By SAS rule)
$$\therefore$$ AB = AD ...(By CPCT)
Also, AB = DC and AD = BC
(opposite sides of parallelogram)
$$\therefore$$ AB = BC = DC = AD ...(i)

Similarly, in $$\triangle{ABC}$$ and $$\triangle{BAD}$$, we have,
AC = BD ...(given)
AB = BA ...(Common side)
BC = AD ...(from(i))
$$\therefore$$ $$\triangle{ABC}$$ $$\displaystyle \cong$$ $$\triangle{BAD}$$
(By SSS rule)
$$\therefore$$ $$\angle{ABC}$$ = $$\angle{BAD}$$ ...(ii)(By CPCT)

But $$\angle{ABC}$$ + $$\angle{BAD}$$ = $$180^\circ$$ >br> (linear pair axiom)
$$\angle{ABC}$$ = $$\angle{BAD}$$ = $$90^\circ$$ ...(from (ii))
$$\therefore$$ AB = BC = CD = DA, and $$\angle{A}$$ = $$90^\circ$$

Hence, it is proved that, ABCD is a square.