Quadrilaterals

Given: ABCD is a parallelogram in which diagonal AC = diagonal BD.

To prove: ABCD is a rectangle.

Proof:

In $\triangle ABC$ and $\triangle DCB$:
- $AB = DC$ (opposite sides of a parallelogram)
- $BC = BC$ (common)
- $AC = DB$ (given, diagonals are equal)

By SSS congruence rule:
$$\triangle ABC \cong \triangle DCB$$

Therefore, $\angle ABC = \angle DCB$ (CPCT)

Since $AB \parallel DC$ and $BC$ is a transversal:
$$\angle ABC + \angle DCB = 180^\circ \quad \text{(co-interior angles)}$$

$$\Rightarrow 2\angle ABC = 180^\circ$$

$$\Rightarrow \angle ABC = 90^\circ$$

Since ABCD is a parallelogram with one angle $= 90^\circ$, ABCD is a rectangle. $\blacksquare$

Given: ABCD is a square, i.e., $AB = BC = CD = DA$ and all angles are $90^\circ$.

To prove: (a) $AC = BD$, (b) diagonals bisect each other, (c) they bisect at right angles.

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(a) AC = BD:

In $\triangle ABC$ and $\triangle BAD$:
- $AB = BA$ (common)
- $BC = AD$ (sides of a square)
- $\angle ABC = \angle BAD = 90^\circ$

By SAS: $\triangle ABC \cong \triangle BAD$

$\therefore AC = BD$ (CPCT) $\checkmark$

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(b) Diagonals bisect each other:

Let diagonals AC and BD intersect at O.

In $\triangle AOB$ and $\triangle COD$:
- $AB = CD$ (sides of a square)
- $\angle OAB = \angle OCD$ (alternate interior angles, $AB \parallel CD$)
- $\angle OBA = \angle ODC$ (alternate interior angles)

By AAS: $\triangle AOB \cong \triangle COD$

$\therefore OA = OC$ and $OB = OD$ (CPCT)

Hence diagonals bisect each other. $\checkmark$

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(c) Diagonals bisect at right angles:

In $\triangle AOB$ and $\triangle AOD$:
- $OB = OD$ (proved above)
- $AB = AD$ (sides of a square)
- $AO = AO$ (common)

By SSS: $\triangle AOB \cong \triangle AOD$

$\therefore \angle AOB = \angle AOD$ (CPCT)

But $\angle AOB + \angle AOD = 180^\circ$ (linear pair)

$$\Rightarrow 2\angle AOB = 180^\circ \Rightarrow \angle AOB = 90^\circ$$

Hence diagonals bisect each other at right angles. $\blacksquare$

Given: ABCD is a parallelogram in which diagonal AC bisects $\angle A$, i.e., $\angle DAC = \angle BAC$.

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**(i) AC bisects $\angle C$:**

Since $AB \parallel DC$ and $AC$ is a transversal:
$$\angle BAC = \angle DCA \quad \text{(alternate interior angles)} \quad \ldots(1)$$

Since $AD \parallel BC$ and $AC$ is a transversal:
$$\angle DAC = \angle BCA \quad \text{(alternate interior angles)} \quad \ldots(2)$$

But $\angle DAC = \angle BAC$ (given) $\ldots(3)$

From (1), (2) and (3):
$$\angle DCA = \angle BCA$$

Therefore, AC bisects $\angle C$ also. $\checkmark$

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(ii) ABCD is a rhombus:

In $\triangle ABC$:
$$\angle BAC = \angle BCA \quad \text{(from (1) and (3))}$$

$\Rightarrow BC = AB$ (sides opposite equal angles are equal)

Since ABCD is a parallelogram, $AB = CD$ and $BC = AD$.

Therefore $AB = BC = CD = DA$.

Hence ABCD is a rhombus. $\blacksquare$

Given: ABCD is a rectangle; AC bisects $\angle A$ and $\angle C$.

So $\angle DAC = \angle BAC = 45^\circ$ and $\angle DCA = \angle BCA = 45^\circ$ (since each angle of a rectangle is $90^\circ$).

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(i) ABCD is a square:

In $\triangle ABC$:
$$\angle BAC = 45^\circ \text{ and } \angle BCA = 45^\circ$$
$$\Rightarrow \angle BAC = \angle BCA$$
$$\Rightarrow BC = AB \quad \text{(sides opposite equal angles)}$$

Since ABCD is a rectangle, $AB = CD$ and $BC = AD$.

Therefore $AB = BC = CD = DA$.

A rectangle with all sides equal is a square. $\checkmark$

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**(ii) BD bisects $\angle B$ and $\angle D$:**

Since ABCD is a square, $AB = BC = CD = DA$.

In $\triangle ABD$:
$$AB = AD \Rightarrow \angle ABD = \angle ADB$$

Also $\angle ABD + \angle ADB = 90^\circ$ (since $\angle A = 90^\circ$)
$$\Rightarrow \angle ABD = \angle ADB = 45^\circ = \frac{1}{2} \times 90^\circ$$

So BD bisects $\angle B$.

Similarly in $\triangle BCD$: $BC = CD \Rightarrow \angle CBD = \angle CDB = 45^\circ$

So BD bisects $\angle D$ also. $\blacksquare$

Given: ABCD is a parallelogram; P and Q are on diagonal BD such that $DP = BQ$.

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**(i) $\triangle APD \cong \triangle CQB$:**

In $\triangle APD$ and $\triangle CQB$:
- $AD = CB$ (opposite sides of parallelogram)
- $DP = BQ$ (given)
- $\angle ADP = \angle CBQ$ (alternate interior angles, since $AD \parallel BC$ and $BD$ is transversal)

By SAS: $\triangle APD \cong \triangle CQB$ $\checkmark$

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**(ii) $AP = CQ$:**

From (i), $AP = CQ$ (CPCT) $\checkmark$

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**(iii) $\triangle AQB \cong \triangle CPD$:**

In $\triangle AQB$ and $\triangle CPD$:
- $AB = CD$ (opposite sides of parallelogram)
- $BQ = DP$ (given)
- $\angle ABQ = \angle CDP$ (alternate interior angles, since $AB \parallel CD$ and $BD$ is transversal)

By SAS: $\triangle AQB \cong \triangle CPD$ $\checkmark$

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**(iv) $AQ = CP$:**

From (iii), $AQ = CP$ (CPCT) $\checkmark$

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(v) APCQ is a parallelogram:

From (ii): $AP = CQ$
From (iv): $AQ = CP$

Since both pairs of opposite sides are equal, APCQ is a parallelogram. $\blacksquare$

Given: ABCD is a parallelogram; $AP \perp BD$ and $CQ \perp BD$.

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**(i) $\triangle APB \cong \triangle CQD$:**

In $\triangle APB$ and $\triangle CQD$:
- $\angle APB = \angle CQD = 90^\circ$ (given, AP and CQ are perpendiculars)
- $AB = CD$ (opposite sides of parallelogram)
- $\angle ABP = \angle CDQ$ (alternate interior angles, since $AB \parallel CD$ and $BD$ is transversal)

By AAS: $\triangle APB \cong \triangle CQD$ $\checkmark$

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**(ii) $AP = CQ$:**

From (i), $AP = CQ$ (CPCT) $\blacksquare$

Given: ABCD is a trapezium with $AB \parallel CD$ and $AD = BC$.

Construction: Extend AB to E and draw $CE \parallel DA$, meeting AB produced at E.

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**(i) $\angle A = \angle B$:**

Since $AD \parallel CE$ and $AE \parallel DC$ (by construction and given), ADCE is a parallelogram.

$\therefore AD = CE$ (opposite sides of parallelogram)

But $AD = BC$ (given), so $BC = CE$.

In $\triangle BCE$: $BC = CE \Rightarrow \angle CBE = \angle CEB$ (base angles of isosceles triangle)

$\angle CBE = \angle ABC$ (same angle)

Now, $\angle DAB + \angle CBE = 180^\circ$ ... wait, let us use co-interior angles.

Since $AD \parallel CE$:
$$\angle DAE + \angle CEA = 180^\circ \quad \text{(co-interior angles on same side of AE)}$$

In $\triangle BCE$, $BC = CE$, so $\angle CBE = \angle CEB$.

$$\angle ABC = 180^\circ - \angle CBE = 180^\circ - \angle CEB$$

Also $\angle DAB + \angle CEB = 180^\circ$ (since $\angle DAE = \angle DAB$ and $\angle CEA = \angle CEB$, co-interior angles with $AD \parallel CE$)

$$\Rightarrow \angle DAB = 180^\circ - \angle CEB = \angle ABC$$

$$\therefore \angle A = \angle B \checkmark$$

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**(ii) $\angle C = \angle D$:**

Since $AB \parallel CD$:
$$\angle A + \angle D = 180^\circ \quad \text{and} \quad \angle B + \angle C = 180^\circ$$

Since $\angle A = \angle B$:
$$\angle D = 180^\circ - \angle A = 180^\circ - \angle B = \angle C$$

$$\therefore \angle C = \angle D \checkmark$$

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**(iii) $\triangle ABC \cong \triangle BAD$:**

In $\triangle ABC$ and $\triangle BAD$:
- $AB = BA$ (common)
- $BC = AD$ (given)
- $\angle ABC = \angle BAD$ (proved in (i))

By SAS: $\triangle ABC \cong \triangle BAD$ $\checkmark$

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(iv) diagonal AC = diagonal BD:

From (iii): $AC = BD$ (CPCT) $\blacksquare$