QUADRILATERALS-Notes
Maths - Notes
QUADRILATERALS
A quadrilateral is a closed figure with four sides, four angles, and four vertices.
Types of Quadrilaterals
- Parallelogram:
Opposite sides are parallel and equal. - Rectangle:
Parallelogram with all angles equal to 90°. - Rhombus:
Parallelogram with all sides equal. - Square:
Parallelogram with all sides equal and each angle 90°. - Trapezium:
Only one pair of opposite sides is parallel.
Properties of a Parallelogram
- Opposite sides are equal and parallel.
- Opposite angles are equal.
- The diagonals bisect each other.
- Each diagonal divides the parallelogram into two congruent triangles.
Conditions for a Quadrilateral to be a Parallelogram
- Both pairs of opposite sides are equal, or
- Both pairs of opposite sides are parallel, or
- One pair of opposite sides is equal and parallel, or
- The diagonals bisect each other.
Theorem-1
Theorem-1: The diagonals bisect each other.
Proof:
In the given parallelogram \(ABCD\):
We have two triangles: \(\triangle ABC\) and \(\triangle CDA\).
Since \(BC \parallel AD\) and \(AC\) is a transversal,
\[
\scriptsize\angle BCA = \angle DAC\text{ (Alternate interior angles)}
\]
Also, \(AB \parallel DC\) and \(AC\) is a transversal,
\[
\scriptsize\angle BAC = \angle DCA\text{ (Alternate interior angles)}
\]
The side \(AC\) is common to both triangles,
\[
\scriptsize AC = AC\text{ (Common side)}
\]
Therefore, by ASA congruence rule,
\[
\triangle ABC \cong \triangle CDA
\]
Hence, the diagonal \(AC\) divides the parallelogram \(ABCD\) into two congruent triangles
\(ABC\) and \(CDA\).
Theorem-2
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-3
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-4
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.
Theorem-5
Theorem: If the diagonals of a quadrilateral bisect each other, then it is a parallelogram.
Given: In quadrilateral \(ABCD\), diagonals \(AC\) and \(BD\) intersect at \(O\) such that \(AO=OC\) and \(BO=OD\).
To prove: \(ABCD\) is a parallelogram.
Proof:
Consider \(\triangle AOB\) and \(\triangle COD\).
\[ \begin{aligned} AO &= OC\text{ (Given)}\\ BO &= OD\text{ (Given)}\\ \angle AOB &= \angle COD \end{aligned} \\\text{ (Vertically opposite angles)} \] \[ \therefore\ \triangle AOB \cong \triangle COD \quad (\text{SAS}) \] By CPCT, \[ AB = CD \tag{1} \]Next, consider \(\triangle AOD\) and \(\triangle COB\).
\[ \begin{aligned} AO &= OC \text{ (Given)}\\ OD &= OB \text{ (Given)}\\ \angle AOD &= \angle COB \end{aligned} \\\text{ (Vertically opposite angles)} \] \[ \therefore\ \triangle AOD \cong \triangle COB \quad (\text{SAS}) \] By CPCT, \[ AD = BC \tag{2} \]From (1) and (2), both pairs of opposite sides of \(ABCD\) are equal. Hence, by the converse criterion for parallelograms (a quadrilateral with both pairs of opposite sides equal is a parallelogram),
\[ \boxed{\,ABCD \text{ is a parallelogram.}\,} \]
Example-1
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.
Example-2
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.
\[ \boxed{AC \perp BD} \]
Example-3
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.
Given:
\[ AB = AC\\ \angle PAD = \angle DAC\\ CD \parallel AB \]To prove:
\[ (i)\ \angle DAC = \angle BCA, \\ (ii)\ ABCD\ \text{is a parallelogram.} \]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)
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.
\[ \therefore\ ABCD\ \text{is a parallelogram.} \] \[ \boxed{ABCD\ \text{is a parallelogram.}} \]
Example-4
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.
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
\(ABCD\).
To prove:
\[ ABCD\ \text{is a rectangle.} \]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.
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 \(90^\circ\).
\[ \boxed{ABCD\ \text{is a rectangle.}} \]
The Mid-point Theorem
The line segment joining the midpoints of two sides of a triangle is parallel to the third side and half of it in length.
Theorem-6
Theorem: The line segment joining the midpoints of any two sides of a triangle is parallel to the third side and is half of it.
Given:
In \(\triangle ABC\), let \(D\) and \(E\) be the midpoints of sides \(AB\) and \(AC\) respectively.
To prove:
\[ DE \parallel BC \text{ and } DE = \frac{1}{2}BC \]Construction:
Join \(D\) and \(E\). Draw a line through \(E\) parallel to \(BC\), intersecting the extension of \(AB\) at \(F\).
Proof:
In \(\triangle AFE\) and \(\triangle CDE\),
\[ AE = CE \text{ (E is the midpoint of AC)} \\ \angle AEF = \angle CED \\\text{(Vertically opposite angles)} \\ AF = CD \\\text{(F and D are midpoints of AB and its extension)} \] \[ \therefore \triangle AFE \cong \triangle CDE \quad (\text{By ASA congruence rule}) \]Hence, the corresponding sides are equal:
\[ EF = DE \]Now, the quadrilateral \(BCDF\) is a parallelogram because its opposite sides are parallel and equal.
\[ \therefore\ BD \parallel CF \quad \text{and} \quad BC = DF \]Since \(E\) and \(D\) are the midpoints,
\[ DE = \frac{1}{2}DF = \frac{1}{2}BC \] \[ \boxed{DE \parallel BC \quad \text{and} \quad DE = \frac{1}{2}BC} \]Hence, the line segment joining the midpoints of two sides of a triangle is parallel to the third side and equal to half of it.
Important Point
- A diagonal of a parallelogram divides it into two congruent triangles.
- In a parallelogram,
- opposite sides are equal
- opposite angles are equal
- diagonals bisect each other
- Diagonals of a rectangle bisect each other and are equal and vice-versa.
- Diagonals of a rhombus bisect each other at right angles and vice-versa.
- Diagonals of a square bisect each other at right angles and are equal, and vice-versa.
- The line-segment joining the mid-points of any two sides of a triangle is parallel to the third side and is half of it.
- A line through the mid-point of a side of a triangle parallel to another side bisects the third side.
