QUADRILATERALS-Notes

Theorem-2: In a parallelogram, opposite sides are equal.

From the previous theorem, we have already proved that

\[ \triangle ABC \cong \triangle CDA \]

By CPCT (Corresponding Parts of Congruent Triangles), we get:

\[ AD = BC \quad \text{and} \quad AB = DC \]

Hence, in a parallelogram, the opposite sides are equal.

Theorem 8.3: If each pair of opposite sides of a quadrilateral is equal, then it is a parallelogram.

Proof:

Let the sides of quadrilateral ABCD be AB, BC, CD, and DA, such that:

\[ AB = CD \quad \text{and} \quad BC = AD \]

Consider triangles \( \triangle ABC \) and \( \triangle CDA \).

We have:

\[ \begin{aligned} AB &= CD &(\text{Given}) \\ BC &= AD &(\text{Given}) \\ AC &= AC &(\text{Common side}) \end{aligned} \]

\[\therefore \, \triangle ABC \cong \triangle CDA \\(\text{By SSS congruence rule})\]

By CPCT (Corresponding Parts of Congruent Triangles):

\[ \begin{aligned} \angle DCA &= \angle CAB, \\ \angle DAC &= \angle BCA. \end{aligned} \]

Now, in quadrilateral ABCD:

\[ \angle BCD = \angle DCA + \angle BCA \] \[ \angle DAB = \angle CAB + \angle DAC \]

Since the corresponding component angles are equal, \[ \angle BCD = \angle DAB. \] Similarly, \[ \angle ABC = \angle ADC. \]

Therefore, in quadrilateral ABCD, both pairs of opposite angles are equal.

Hence, quadrilateral \(ABCD\) is a parallelogram, since its opposite sides and opposite angles are equal.

Theorem 8.4: In a parallelogram, opposite angles are equal.

\(\textbf{Proof:}\) Let ABCD be a parallelogram. Draw diagonal AC. Since \(AB \parallel CD\) and AC is a transversal, \[\small \angle BAC = \angle DCA \text{ (Alternate interior angles)} \] Also, \(AD \parallel BC\) and AC is a transversal, \[\small \angle DAC = \angle BCA \text{ (Alternate interior angles)} \]

In \(\triangle ABC\) and \(\triangle CDA\)

\[ AB = CD\\ AC = AC\\ \angle BAC = \angle DCA \] \[ \therefore\triangle ABC \cong \triangle CDA \ (\text{ASA rule}) \] Hence, \[ \left. \begin{array}{l} \angle ABC = \angle CDA \\ \angle BCD = \angle DAB \end{array} \right\} \text{ (CPCT)} \] Therefore, the opposite angles of a parallelogram are equal.

Question: Show that each angle of a rectangle is a right angle.

Given: Let \(ABCD\) be a rectangle in which one angle, \(\angle A = 90^\circ.\)

To prove: Each angle of the rectangle is a right angle.

Proof:

In a rectangle \(ABCD\), \(AD \parallel BC\) and \(AB\) is a transversal.

The sum of the interior angles on the same side of a transversal is \(180^\circ\). Therefore,

\[ \begin{aligned} \angle A + \angle B &= 180^\circ \\ 90^\circ + \angle B &= 180^\circ \\ \Rightarrow \angle B &= 180^\circ - 90^\circ = 90^\circ \end{aligned} \]

Similarly, \(AB \parallel CD\) and \(BC\) is a transversal.

\[ \begin{aligned} \angle B + \angle C &= 180^\circ \\ \Rightarrow 90^\circ + \angle C &= 180^\circ \\ \Rightarrow \angle C &= 90^\circ \end{aligned} \]

Also, \(AD \parallel BC\) and \(DC\) is a transversal.

\[ \begin{aligned} \angle C + \angle D &= 180^\circ \\ \Rightarrow 90^\circ + \angle D &= 180^\circ \\ \Rightarrow \angle D &= 90^\circ \end{aligned} \]

Hence, all four angles of the rectangle \(ABCD\) are right angles.

\[ \boxed{\angle A = \angle B = \angle C = \angle D = 90^\circ} \]

Therefore, each angle of a rectangle is a right angle.

Question: Show that the diagonals of a rhombus are perpendicular to each other.

Given: Let \(ABCD\) be a rhombus in which

\[ AB = BC = CD = DA \\ \text{(All sides of a rhombus are equal).} \]

To prove: The diagonals of the rhombus are perpendicular to each other.

Construction: Draw diagonals \(AC\) and \(BD\) intersecting each other at \(O\).

Proof:

Since the diagonals of a parallelogram bisect each other, in rhombus \(ABCD\) (which is a special type of parallelogram), we have:

\[ AO = CO \quad \text{and} \quad BO = DO \]

Consider triangles \( \triangle AOD \) and \( \triangle COD \).

\[ \begin{aligned} AO &= CO\text{ (Diagonals bisect each other)} \\ OD &= OD\text{ (Common side)} \\ AD &= CD \end{aligned} \\\text{ (All sides of a rhombus are equal)} \] \[ \therefore \, \triangle AOD \cong \triangle COD \\ (\text{By SSS congruence rule}) \]

By CPCT (Corresponding Parts of Congruent Triangles):

\[ \angle AOD = \angle COD \]

Also, \(\angle AOD\) and \(\angle COD\) form a linear pair.

\[ \angle AOD + \angle COD = 180^\circ \] \[ \Rightarrow \angle AOD + \angle AOD = 180^\circ \] \[ \Rightarrow 2\angle AOD = 180^\circ \] \[ \Rightarrow \angle AOD = 90^\circ \]

Hence, the diagonals of a rhombus are perpendicular to each other.

\[ \boxed{AC \perp BD} \]

Question: In an isosceles triangle \(ABC\), \(AB = AC\). The ray \(AD\) bisects the exterior angle \(PAC\) and \(CD \parallel AB\) (see figure). Show that:

(i) \(\angle DAC = \angle BCA\)
(ii) \(ABCD\) is a parallelogram.

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Given:

\[ AB = AC\\ \angle PAD = \angle DAC\\ CD \parallel AB \]

To prove:

\[ (i)\ \angle DAC = \angle BCA, \\ (ii)\ ABCD\ \text{is a parallelogram.} \]

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Proof:

Since \(AB = AC\), \(\triangle ABC\) is an isosceles triangle. Therefore,

\[ \angle ABC = \angle BCA \]

Now, since \(AD\) bisects the exterior angle \(PAC\),

\[ \angle PAD = \angle DAC \]

In \(\triangle ABC\), \(\angle PAC\) is an exterior angle. By the Exterior Angle Theorem:

\[ \angle PAC = \angle ABC + \angle BCA \]

But, since \(\triangle ABC\) is isosceles,

\[ \angle ABC = \angle BCA \] \[ \Rightarrow \angle PAC = 2\angle BCA \tag{1} \]

Also, since \(AD\) bisects \(\angle PAC\),

\[ \angle PAC = 2\angle DAC \tag{2} \]

From equations (1) and (2),

\[ 2\angle BCA = 2\angle DAC \] \[ \Rightarrow \angle BCA = \angle DAC \]

Hence, \(\angle BCA = \angle DAC\). (Proved)

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To prove: \(ABCD\) is a parallelogram.

In quadrilateral \(ABCD\), we have:

\[ \angle BCA = \angle DAC \quad \text{(proved above)} \]

and

\[ AB \parallel CD \quad \text{(given)} \]

Since one pair of opposite sides of a quadrilateral are parallel and the included angles are equal, the other pair of sides \(BC\) and \(AD\) are also parallel.

\[ \therefore\ ABCD\ \text{is a parallelogram.} \] \[ \boxed{ABCD\ \text{is a parallelogram.}} \]

Question: Two parallel lines \(l\) and \(m\) are intersected by a transversal \(p\) (see figure). Show that the quadrilateral formed by the bisectors of the interior angles is a rectangle.

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Given:

\[ l \parallel m \]

and transversal \(p\) cuts them at points \(A\) and \(C\).
The bisectors of the interior angles at \(A\) and \(C\) intersect each other to form quadrilateral \(ABCD\).

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To prove:

\[ ABCD\ \text{is a rectangle.} \]

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Proof:

Since \(l \parallel m\) and \(p\) is a transversal,

\[ \angle PAC = \angle ACR \\\text{ (Alternate interior angles)} \]

Now, let the bisectors of \(\angle PAC\) and \(\angle ACR\) meet the transversal \(AC\) at points \(B\) and \(D\) respectively.

Therefore,

\[ \frac{1}{2}\angle PAC = \frac{1}{2}\angle ACR \] \[ \Rightarrow \angle BAC = \angle ACD \]

Hence, the lines \(AB\) and \(CD\) are parallel because alternate angles are equal.

\[ \therefore AB \parallel CD \tag{1} \]

Similarly, since the bisectors of the angles on the other side of \(AC\), say \(\angle SAC\) and \(\angle ACQ\), are equal, we have:

\[ \angle SAC = \angle ACQ \\\text{ (Alternate interior angles)} \] \[ \frac{1}{2}\angle SAC = \frac{1}{2}\angle ACQ \] \[ \Rightarrow \angle CAD = \angle ACB \]

Hence, \(AD \parallel BC\).

\[ \therefore AD \parallel BC \tag{2} \]

From (1) and (2), we have both pairs of opposite sides of \(ABCD\) parallel. Therefore, \(ABCD\) is a parallelogram.

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Now, to show that \(ABCD\) is a rectangle:

At point \(A\), \[ \angle PAC + \angle SAC = 180^\circ \\ \text{(Linear pair on line } l) \] Dividing both sides by 2, \[ \frac{1}{2}(\angle PAC + \angle SAC) = 90^\circ \] \[ \Rightarrow \angle BAC + \angle CAD = 90^\circ \] \[ \Rightarrow \angle BAD = 90^\circ \]

Since one angle of the parallelogram is \(90^\circ\), all its angles are \(90^\circ\).

\[ \boxed{ABCD\ \text{is a rectangle.}} \]