Solution:

Given:

\(ABCD\) is a square with
\[ AB = BC = CD = DA \]

To Prove:

\[ AC = BD\\AO = OC\\BO = OD \] Diagonals bisect each other at \(90^\circ\).

Proof:

Step 1: Diagonals are equal. Since all sides of the square are equal and each angle is \(90^\circ\), consider triangle \(ABC\): \[ AC^2 = AB^2 + BC^2 \] If \(AB = BC = a\): \[\begin{aligned} AC^2 &= a^2 + a^2 \\&= 2a^2 \\\implies AC &= \sqrt{2}a \end{aligned}\] Similarly, \(BD = \sqrt{2}a\). Thus, the diagonals are equal: \[ AC = BD \] Step 2: Diagonals bisect each other. In any square, the diagonals bisect each other, so \[ AO = OC,\quad BO = OD \] Step 3: Diagonals intersect at right angles.
Consider \(\triangle AOD\) and \(\triangle COD\): \[\begin{aligned}AD &= CD\text{ (sides of square)}\\ OD &= OD\text{ (common side)}\\ AO &= CO\text{ (half of equal diagonals)}\end{aligned} \] \[ \triangle AOD \cong \triangle COD\\\text{By SSS-Rule} \] So, \[ \angle AOD = \angle COD \] Since these angles form a straight line along diagonal \(AC\), \[ \angle AOD + \angle COD = 180^\circ \] Therefore, \[\begin{aligned} 2\angle AOD &= 180^\circ \\\implies \angle AOD &= 90^\circ \end{aligned}\] Diagonals bisect each other at \(90^\circ\).

Conclusion:

The diagonals of a square are equal, bisect each other, and intersect at \(90^\circ\).

Solution:
\(\textbf{Given:}\\\\\) \(ABCD\) is a parallelogram in which the diagonal \(AC\) bisects \(\angle A\). \[ \begin{aligned} AB &= CD, \\ BC &= AD, \\ \angle DAC &= \angle CAB. \end{aligned} \] \(\textbf{To prove:}\) \[ \text{(i) } AC \text{ bisects } \angle C, \\ \text{(ii) } ABCD \text{ is a rhombus.} \] \(\textbf{Proof:}\)

Consider triangles \( \triangle ACD \) and \( \triangle ABC \). \[ AB = CD \\ \small\text{(Opposite sides of a parallelogram)} \\ BC = AD \\ \small\text{(Opposite sides of a parallelogram)} \\ AC = AC \\ \small\text{(Common side)} \] \[\therefore \triangle ACD \cong \triangle ABC \\ \small(\text{By SSS congruence})\] By \(CPCT\) (Corresponding Parts of Congruent Triangles), \[ \angle DCA = \angle BCA. \] Thus, the diagonal \(AC\) bisects \(\angle C\). \[ \text{Hence, (i) is proved.} \] Now, since \(AC\) bisects \(\angle A\), \[ \angle DAC = \angle CAB. \] From the congruence of triangles
\( \triangle ACD \) and \( \triangle ABC \), \[ \begin{aligned} AD &= DC \quad \text{and} \\ AB &= BC. \end{aligned} \] \(\therefore AB = BC = CD = DA.\) \[ \boxed{\text{Hence, } ABCD \text{ is a rhombus.}} \]

Solution:

Given: ABCD is a rectangle in which the diagonal \(AC\) bisects both \(\angle A\) and \(\angle C\).

To prove:

(i) ABCD is a square

(ii) Diagonal \(BD\) bisects \(\angle B\) as well as \(\angle D\)

Proof:

Since ABCD is a rectangle, all its angles are right angles, that is,

\(\angle A = \angle B = \angle C = \angle D = 90^{\circ}\)

Because \(AC\) bisects \(\angle A\), we have

\(\angle CAB = \angle DAC = \frac{1}{2} \times 90^{\circ} = 45^{\circ}\)

In triangle \(\triangle ABC\),

\(\angle ABC = 90^{\circ}\) and \(\angle CAB = 45^{\circ}\).

Therefore, the remaining angle \(\angle ACB = 45^{\circ}\).

Hence, triangle \(\triangle ABC\) is a right isosceles triangle, and so

\[AB = BC\tag{1}\]

Similarly, since \(AC\) bisects \(\angle C\),

\(\angle BCA = \angle ACD = 45^{\circ}\).

In triangle \(\triangle BCD\), which is right-angled at C, we have

\(\angle BCD = 90^{\circ}\) and \(\angle BCA = 45^{\circ}\).

Therefore, triangle \(\triangle BCD\) is also right isosceles, and hence

\[BC = CD\tag{2}\]

From (1) and (2), we get

\(AB = BC = CD\).

Since opposite sides of a rectangle are equal, \(AD = BC\).

Therefore, all sides are equal:

\(AB = BC = CD = DA\).

Hence, ABCD is a square. (Part \((i)\) proved)

---

(ii) To prove: Diagonal \(BD\) bisects \(\angle B\) and \(\angle D\).

In square ABCD, consider triangles \(\triangle ABD\) and \(\triangle CBD\).

We have:

\[\scriptsize\begin{aligned} AB &= CB\text{ (all sides of a square are equal)}\\ AD &= CD\text{ (all sides of a square are equal)}\\ BD &= BD\text{ (common side)} \end{aligned}\] Therefore, \[\triangle ABD \cong \triangle CBD\\\text{ (by SSS congruence)}\] By CPCT, \[\angle ABD = \angle DBC\]

Hence, diagonal \(BD\) bisects \(\angle B\).

Similarly, it can be shown that \(BD\) also bisects \(\angle D\).

Therefore, diagonal \(BD\) bisects both \(\angle B\) and \(\angle D\).

Hence proved.

Solution:

Given:

\[ \begin{aligned} DP &= BQ,\\ AD &= BC,\\ AB &= CD \end{aligned} \]

To Prove: (i) to (v) as stated above.

---

Proof:

(i) To prove

\(\triangle APD \cong \triangle CQB\)

In triangles \(APD\) and \(CQB\):

\[ \begin{aligned} DP &= BQ \text{ (Given)}\\ \angle APD &= \angle CBQ\; (\text{Alternate angles, since } \\AD &\parallel BC\; BD \text{ is a transversal})\\ AD &= BC \text{ (Given)} \end{aligned} \] \[ \therefore \triangle APD \cong \triangle CQB \quad (\text{By SAS rule}) \] \[ \Rightarrow AP = CQ \quad (\text{By CPCT}) \]

Hence proved (i) and (ii).

---

(iii) To prove:

\(\triangle AQB \cong \triangle CPD\)

In triangles \(AQB\) and \(CPD\):

\[ \begin{aligned} BQ &= DP \text{ (Given)}\\ \angle ABQ &= \angle CDP \end{aligned}\]

(Alternate angles, since \(AB \parallel CD\) and \(BD\) is a transversal)

\[ AB = CD \]

(Opposite sides of a parallelogram are equal)

\[ \therefore \triangle AQB \cong \triangle CPD \\(\text{By SAS rule}) \] \[ \Rightarrow AQ = CP \quad (\text{By CPCT}) \]

Hence proved (iii) and (iv).

---

(v) To prove

\(APCQ\) is a parallelogram

From the above results, we have

\[ \begin{aligned} AP &= CQ,\\ AQ &= CP. \end{aligned} \]

When both pairs of opposite sides of a quadrilateral are equal, the quadrilateral is a parallelogram.

\[ \therefore APCQ \text{ is a parallelogram.} \]

Hence proved.