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Answer :

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.

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- Show that if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus.
- Show that the diagonals of a square are equal and bisect each other at right angles.
- Diagonal AC of a parallelogram ABCD bisects \(\angle{A}\) (see figure). Show that (i) It bisects \(\angle{C}\) also, (ii) ABCD is a rhombus.
- ABCD is a rhombus. Show that diagonal AC bisects \(\angle{A}\) as well as \(\angle{C}\) and diagonal BD bisects \(\angle{B}\) as well as \(\angle{D}\).
- ABCD is a rectangle in which diagonal AC bisects \(\angle{A}\) as well as \(\angle{C}\). Show that: (i) ABCD is a square (ii) Diagonal BD bisects \(\angle{B}\) as well as \(\angle{D}\).
- In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP = BQ (see figure). Show thati) \(\triangle{APD}\) \(\displaystyle \cong \) \(\triangle{CQB}\) ii) AP = CQiii) \(\triangle{AQB}\) \(\displaystyle \cong \) \(\triangle{CPD}\) iv) AQ = CPv) APCQ is a parallelogram.
- ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD (see figure). Show that (i) \(\triangle{APB}\) \(\displaystyle \cong \) \(\triangle{CQD}\)(ii) AP = CQ
- In \(\triangle{ABC}\) and \(\triangle{DEF}\), AB = DE, AB || DE, BC = EF and BC || EF. Vertices A, B and C are joined to vertices D, E and F, respectively (see figure). Show that (i) Quadrilateral ABED is a parallelogram (ii) Quadrilateral BEFC is a parallelogram (iii) AD || CF and AD = CF (iv) Quadrilateral ACFD is a Parallelogram (v) AC = DF (vi) \(\triangle{ABC}\) \(\displaystyle \cong \) \(\triangle{DEF}\)
- ABCD is a trapezium in which AB || CD and AD = BC (see figure). Show that i) \(\angle{A}\) = \(\angle{B}\) ii) \(\angle{C}\) = \(\angle{D}\) iii) \(\triangle{ABC}\) \(\displaystyle \cong \) \(\triangle{BAD}\) iv) Diagonal AC = Diagonal BD [Hint: Extend AB and draw a line through C parallel to DA intersecting AB produced at E].

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