Chapter 8: Quadrilaterals
Exercise 8.1
Q1
If the diagonals of a parallelogram are equal, then show that it is a rectangle.
Solution:
Let ABCD be a parallelogram. To show that ABCD is a rectangle, we have to prove that one of its interior angles is \( 90^{\circ} \).
In \( \Delta ABC \) and \( \Delta DCB \):
\[ AB = DC \quad \text{(Opposite sides of a parallelogram are equal)} \]
\[ BC = BC \quad \text{(Common)} \]
\[ AC = DB \quad \text{(Given)} \]
\[ \Rightarrow \Delta ABC \cong \Delta DCB \quad \text{(By SSS Congruence rule)} \]
\[ \Rightarrow \angle ABC = \angle DCB \]
It is known that the sum of the measures of angles on the same side of transversal is \( 180^{\circ} \).
\[ \angle ABC + \angle DCB = 180^{\circ} \quad \text{(Since } AB \parallel CD \text{)} \]
\[ \Rightarrow \angle ABC + \angle ABC = 180^{\circ} \]
\[ \Rightarrow 2\angle ABC = 180^{\circ} \]
\[ \Rightarrow \angle ABC = 90^{\circ} \]
Since ABCD is a parallelogram and one of its interior angles is \( 90^{\circ} \), ABCD is a rectangle.
Q2
Show that if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus.
Solution:
Let ABCD be a quadrilateral, whose diagonals AC and BD bisect each other at right angle i.e., \( OA = OC \), \( OB = OD \), and \( \angle AOB = \angle BOC = \angle COD = \angle AOD = 90^{\circ} \). To prove ABCD a rhombus, we have to prove ABCD is a parallelogram and all the sides of ABCD are equal.
In \( \Delta AOD \) and \( \Delta COD \):
\[ OA = OC \quad \text{(Diagonals bisect each other)} \]
\[ \angle AOD = \angle COD \quad \text{(Given)} \]
\[ OD = OD \quad \text{(Common)} \]
\[ \Rightarrow \Delta AOD \cong \Delta COD \quad \text{(By SAS congruence rule)} \]
\[ \Rightarrow AD = CD \quad \dots(1) \]
Similarly, it can be proved that:
\[ AD = AB \quad \text{and} \quad CD = BC \quad \dots(2) \]
From equations (1) and (2):
\[ AB = BC = CD = AD \]
Since opposite sides of quadrilateral ABCD are equal, it can be said that ABCD is a parallelogram. Since all sides of a parallelogram ABCD are equal, it can be said that ABCD is a rhombus.
Q3
Show that if the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square.
Solution:
Let us consider a quadrilateral ABCD in which the diagonals AC and BD intersect each other at O. It is given that \( AC = BD \), \( OA = OC \), \( OB = OD \), and \( \angle AOB = \angle BOC = \angle COD = \angle AOD = 90^{\circ} \).
To prove ABCD is a square, we have to prove that ABCD is a parallelogram, \( AB = BC = CD = AD \), and one of its interior angles is \( 90^{\circ} \).
In \( \Delta AOB \) and \( \Delta COD \):
\[ AO = CO \quad \text{(Diagonals bisect each other)} \]
\[ OB = OD \quad \text{(Diagonals bisect each other)} \]
\[ \angle AOB = \angle COD \quad \text{(Vertically opposite angles)} \]
\[ \Rightarrow \Delta AOB \cong \Delta COD \quad \text{(SAS congruence rule)} \]
\[ \Rightarrow AB = CD \quad \text{(By CPCT)} \quad \dots(1) \]
\[ \text{And, } \angle OAB = \angle OCD \quad \text{(By CPCT)} \]
However, these are alternate interior angles for line AB and CD and alternate interior angles are equal to each other only when the two lines are parallel.
\[ \Rightarrow AB \parallel CD \quad \dots(2) \]
From equations (1) and (2), ABCD is a parallelogram.
In \( \Delta AOD \) and \( \Delta COD \):
\[ AO = CO \quad \text{(Diagonals bisect each other)} \]
\[ \angle AOD = \angle COD \quad \text{(Given that each is } 90^{\circ} \text{)} \]
\[ OD = OD \quad \text{(Common)} \]
\[ \Rightarrow \Delta AOD \cong \Delta COD \quad \text{(SAS congruence rule)} \]
\[ \Rightarrow AD = DC \quad \dots(3) \]
However, \( AD = BC \) and \( AB = CD \) (Opposite sides of parallelogram ABCD).
\[ \Rightarrow AB = BC = CD = DA \]
Therefore, all the sides of quadrilateral ABCD are equal to each other.
In \( \Delta ADC \) and \( \Delta BCD \):
\[ AD = BC \quad \text{(Already proved)} \]
\[ AC = BD \quad \text{(Given)} \]
\[ DC = CD \quad \text{(Common)} \]
\[ \Rightarrow \Delta ADC \cong \Delta BCD \quad \text{(SSS Congruence rule)} \]
\[ \Rightarrow \angle ADC = \angle BCD \quad \text{(By CPCT)} \]
However, \( \angle ADC + \angle BCD = 180^{\circ} \) (Co-interior angles).
\[ \Rightarrow \angle ADC + \angle ADC = 180^{\circ} \]
\[ \Rightarrow 2\angle ADC = 180^{\circ} \]
\[ \Rightarrow \angle ADC = 90^{\circ} \]
One of the interior angles of quadrilateral ABCD is a right angle. Thus, ABCD is a square.
Q4
Diagonal AC of a parallelogram ABCD bisects \( \angle A \) (see the given figure). Show that:
(i) It bisects \( \angle C \) also,
(ii) ABCD is a rhombus.
(i) It bisects \( \angle C \) also,
(ii) ABCD is a rhombus.
Solution:
(i) ABCD is a parallelogram.
\[ \angle DAC = \angle BCA \quad \text{(Alternate interior angles) ... (1)} \]
\[ \angle BAC = \angle DCA \quad \text{(Alternate interior angles) ... (2)} \]
However, it is given that AC bisects \( \angle A \).
\[ \angle DAC = \angle BAC \quad \dots(3) \]
From equations (1), (2), and (3), we obtain:
\[ \angle DAC = \angle BCA = \angle BAC = \angle DCA \quad \dots(4) \]
\[ \Rightarrow \angle DCA = \angle BCA \]
Hence, AC bisects \( \angle C \).
(ii) From equation (4), we obtain:
\[ \angle DAC = \angle DCA \]
\[ \Rightarrow DA = DC \quad \text{(Side opposite to equal angles are equal)} \]
However, \( DA = BC \) and \( AB = CD \) (Opposite sides of a parallelogram).
\[ \Rightarrow AB = BC = CD = DA \]
Hence, ABCD is a rhombus.
Q5
In parallelogram ABCD, two points P and Q are taken on diagonal BD such that \( DP = BQ \) (see the given figure). Show that:
(i) \( \Delta APD \cong \Delta CQB \)
(ii) \( AP = CQ \)
(iii) \( \Delta AQB \cong \Delta CPD \)
(iv) \( AQ = CP \)
(v) APCQ is a parallelogram
(i) \( \Delta APD \cong \Delta CQB \)
(ii) \( AP = CQ \)
(iii) \( \Delta AQB \cong \Delta CPD \)
(iv) \( AQ = CP \)
(v) APCQ is a parallelogram
Solution:
(i) In \( \Delta APD \) and \( \Delta CQB \):
\[ \angle ADP = \angle CBQ \quad \text{(Alternate interior angles for } BC \parallel AD \text{)} \]
\[ AD = CB \quad \text{(Opposite sides of parallelogram ABCD)} \]
\[ DP = BQ \quad \text{(Given)} \]
\[ \Rightarrow \Delta APD \cong \Delta CQB \quad \text{(Using SAS congruence rule)} \]
(ii) As we observed that \( \Delta APD \cong \Delta CQB \):
\[ AP = CQ \quad \text{(CPCT)} \]
(iii) In \( \Delta AQB \) and \( \Delta CPD \):
\[ \angle ABQ = \angle CDP \quad \text{(Alternate interior angles for } AB \parallel CD \text{)} \]
\[ AB = CD \quad \text{(Opposite sides of parallelogram ABCD)} \]
\[ BQ = DP \quad \text{(Given)} \]
\[ \Rightarrow \Delta AQB \cong \Delta CPD \quad \text{(Using SAS congruence rule)} \]
(iv) As we observed that \( \Delta AQB \cong \Delta CPD \):
\[ AQ = CP \quad \text{(CPCT)} \]
(v) From the result obtained in (ii) and (iv):
\[ AQ = CP \quad \text{and} \quad AP = CQ \]
Since opposite sides in quadrilateral APCQ are equal to each other, APCQ is a parallelogram.
Q6
ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD (See the given figure). Show that:
(i) \( \Delta APB \cong \Delta CQD \)
(ii) \( AP = CQ \)
(i) \( \Delta APB \cong \Delta CQD \)
(ii) \( AP = CQ \)
Solution:
(i) In \( \Delta APB \) and \( \Delta CQD \):
\[ \angle APB = \angle CQD \quad \text{(Each } 90^{\circ} \text{)} \]
\[ AB = CD \quad \text{(Opposite sides of parallelogram ABCD)} \]
\[ \angle ABP = \angle CDQ \quad \text{(Alternate interior angles for } AB \parallel CD \text{)} \]
\[ \Rightarrow \Delta APB \cong \Delta CQD \quad \text{(By AAS congruency)} \]
(ii) By using the above result:
\[ AP = CQ \quad \text{(By CPCT)} \]
Q7
ABCD is a trapezium in which \( AB \parallel CD \) and \( AD = BC \) (see the given figure). Show that:
(i) \( \angle A = \angle B \)
(ii) \( \angle C = \angle D \)
(iii) \( \Delta ABC \cong \Delta BAD \)
(iv) diagonal \( AC = \) diagonal BD
(i) \( \angle A = \angle B \)
(ii) \( \angle C = \angle D \)
(iii) \( \Delta ABC \cong \Delta BAD \)
(iv) diagonal \( AC = \) diagonal BD
Solution:
Let us extend AB. Then, draw a line through C, which is parallel to AD, intersecting AE at point E. It is clear that AECD is a parallelogram.
(i) \( AD = CE \) (Opposite sides of parallelogram AECD).
However, \( AD = BC \) (Given). Therefore, \( BC = CE \).
\[ \angle CEB = \angle CBE \quad \text{(Angle opposite to equal sides are also equal)} \]
Consider parallel lines AD and CE. AE is the transversal line for them.
\[ \angle A + \angle CEB = 180^{\circ} \quad \text{(Angles on the same side of transversal)} \]
\[ \angle A + \angle CBE = 180^{\circ} \quad \text{(Using the relation } \angle CEB = \angle CBE \text{) ... (1)} \]
However, \( \angle B + \angle CBE = 180^{\circ} \) (Linear pair angles) ... (2)
From equations (1) and (2), we obtain: \( \angle A = \angle B \).
(ii) \( AB \parallel CD \).
\[ \angle A + \angle D = 180^{\circ} \quad \text{(Angles on the same side of the transversal)} \]
Also, \( \angle C + \angle B = 180^{\circ} \) (Angles on the same side of the transversal).
\[ \Rightarrow \angle A + \angle D = \angle C + \angle B \]
However, \( \angle A = \angle B \). Therefore, \( \angle C = \angle D \).
(iii) In \( \Delta ABC \) and \( \Delta BAD \):
\[ AB = BA \quad \text{(Common side)} \]
\[ BC = AD \quad \text{(Given)} \]
\[ \angle B = \angle A \quad \text{(Proved before)} \]
\[ \Rightarrow \Delta ABC \cong \Delta BAD \quad \text{(SAS congruence rule)} \]
(iv) We had observed that \( \Delta ABC \cong \Delta BAD \).
\[ \Rightarrow AC = BD \quad \text{(By CPCT)} \]
Exercise 8.2
Q1
ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA. AC is a diagonal. Show that:
(i) \( SR \parallel AC \) and \( SR = \frac{1}{2}AC \)
(ii) \( PQ = SR \)
(iii) PQRS is a parallelogram.
(i) \( SR \parallel AC \) and \( SR = \frac{1}{2}AC \)
(ii) \( PQ = SR \)
(iii) PQRS is a parallelogram.
Solution:
(i) In \( \Delta ADC \), S and R are the mid-points of sides AD and CD respectively. In a triangle, the line segment joining the mid-points of any two sides of the triangle is parallel to the third side and is half of it.
\[ \Rightarrow SR \parallel AC \quad \text{and} \quad SR = \frac{1}{2}AC \quad \dots(1) \]
(ii) In \( \Delta ABC \), P and Q are mid-points of sides AB and BC respectively. Therefore, by using mid-point theorem:
\[ PQ \parallel AC \quad \text{and} \quad PQ = \frac{1}{2}AC \quad \dots(2) \]
Using equations (1) and (2), we obtain:
\[ PQ \parallel SR \quad \text{and} \quad PQ = SR \]
(iii) From the above results, clearly, one pair of opposite sides of quadrilateral PQRS is parallel and equal.
Hence, PQRS is a parallelogram.
Q2
ABCD is a rhombus and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rectangle.
Solution:
In \( \Delta ABC \), P and Q are the mid-points of sides AB and BC respectively.
\[ \Rightarrow PQ \parallel AC \quad \text{and} \quad PQ = \frac{1}{2}AC \quad \text{(Using mid-point theorem) ... (1)} \]
In \( \Delta ADC \), R and S are the mid-points of CD and AD respectively.
\[ \Rightarrow RS \parallel AC \quad \text{and} \quad RS = \frac{1}{2}AC \quad \text{(Using mid-point theorem) ... (2)} \]
From equations (1) and (2), we obtain: \( PQ \parallel RS \) and \( PQ = RS \).
Since in quadrilateral PQRS, one pair of opposite sides is equal and parallel to each other, it is a parallelogram.
Let the diagonals of rhombus ABCD intersect each other at point O. In quadrilateral OMQN:
\[ MQ \parallel ON \quad \text{(Since } PQ \parallel AC \text{)} \]
\[ QN \parallel OM \quad \text{(Since } QR \parallel BD \text{)} \]
Therefore, OMQN is a parallelogram.
\[ \Rightarrow \angle MQN = \angle NOM \]
\[ \Rightarrow \angle PQR = \angle NOM \]
However, \( \angle NOM = 90^{\circ} \) (Diagonals of a rhombus are perpendicular to each other).
\[ \Rightarrow \angle PQR = 90^{\circ} \]
Clearly, PQRS is a parallelogram having one of its interior angles as \( 90^{\circ} \). Hence, PQRS is a rectangle.
Q3
ABCD is a rectangle and P, Q, R and S are mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rhombus.
Solution:
Let us join AC and BD. In \( \Delta ABC \), P and Q are the mid-points of AB and BC respectively.
\[ PQ \parallel AC \quad \text{and} \quad PQ = \frac{1}{2}AC \quad \text{(Mid-point theorem) ... (1)} \]
Similarly in \( \Delta ADC \):
\[ SR \parallel AC \quad \text{and} \quad SR = \frac{1}{2}AC \quad \text{(Mid-point theorem) ... (2)} \]
Clearly, \( PQ \parallel SR \) and \( PQ = SR \). Since in quadrilateral PQRS, one pair of opposite sides is equal and parallel to each other, it is a parallelogram.
\[ PS \parallel QR \quad \text{and} \quad PS = QR \quad \text{(Opposite sides of parallelogram) ... (3)} \]
In \( \Delta BCD \), Q and R are the mid-points of side BC and CD respectively.
\[ \Rightarrow QR \parallel BD \quad \text{and} \quad QR = \frac{1}{2}BD \quad \text{(Mid-point theorem) ... (4)} \]
However, the diagonals of a rectangle are equal. \( AC = BD \) ... (5)
By using equation (1), (2), (3), (4), and (5), we obtain:
\[ PQ = QR = SR = PS \]
Therefore, PQRS is a rhombus.
Q4
ABCD is a trapezium in which \( AB \parallel DC \), BD is a diagonal and E is the mid-point of AD. A line is drawn through E parallel to AB intersecting BC at F (see the given figure). Show that F is the mid-point of BC.
Solution:
Let EF intersect DB at G. By converse of mid-point theorem, we know that a line drawn through the mid-point of any side of a triangle and parallel to another side, bisects the third side.
In \( \Delta ABD \), \( EF \parallel AB \) and E is the mid-point of AD. Therefore, G will be the mid-point of DB.
As \( EF \parallel AB \) and \( AB \parallel CD \):
\[ \Rightarrow EF \parallel CD \quad \text{(Two lines parallel to the same line are parallel to each other)} \]
In \( \Delta BCD \), \( GF \parallel CD \) and G is the mid-point of line BD. Therefore, by using converse of mid-point theorem, F is the mid-point of BC.
Q5
In a parallelogram ABCD, E and F are the mid-points of sides AB and CD respectively (see the given figure). Show that the line segments AF and EC trisect the diagonal BD.
Solution:
ABCD is a parallelogram. \( AB \parallel CD \). And hence, \( AE \parallel FC \).
Again, \( AB = CD \) (Opposite sides of parallelogram ABCD).
\[ \frac{1}{2}AB = \frac{1}{2}CD \]
\[ \Rightarrow AE = FC \quad \text{(E and F are mid-points of side AB and CD)} \]
In quadrilateral AECF, one pair of opposite sides (AE and CF) is parallel and equal to each other. Therefore, AECF is a parallelogram.
\[ \Rightarrow AF \parallel EC \quad \text{(Opposite sides of a parallelogram)} \]
In \( \Delta DQC \), F is the mid-point of side DC and \( FP \parallel CQ \) (as \( AF \parallel EC \)). Therefore, by using the converse of mid-point theorem, it can be said that P is the mid-point of DQ.
\[ \Rightarrow DP = PQ \quad \dots(1) \]
Similarly, in \( \Delta APB \), E is the mid-point of side AB and \( EQ \parallel AP \) (as \( AF \parallel EC \)). Therefore, by using the converse of mid-point theorem, it can be said that Q is the mid-point of PB.
\[ \Rightarrow PQ = QB \quad \dots(2) \]
From equations (1) and (2):
\[ DP = PQ = BQ \]
Hence, the line segments AF and EC trisect the diagonal BD.
Q6
ABC is a triangle right angled at C. A line through the mid-point M of hypotenuse AB and parallel to BC intersects AC at D. Show that:
(i) D is the mid-point of AC
(ii) \( MD \perp AC \)
(iii) \( CM = MA = \frac{1}{2}AB \)
(i) D is the mid-point of AC
(ii) \( MD \perp AC \)
(iii) \( CM = MA = \frac{1}{2}AB \)
Solution:
(i) In \( \Delta ABC \), it is given that M is the mid-point of AB and \( MD \parallel BC \).
Therefore, D is the mid-point of AC. (Converse of mid-point theorem)
(ii) As \( DM \parallel CB \) and AC is a transversal line for them, therefore:
\[ \angle MDC + \angle DCB = 180^{\circ} \quad \text{(Co-interior angles)} \]
\[ \angle MDC + 90^{\circ} = 180^{\circ} \]
\[ \angle MDC = 90^{\circ} \]
Therefore, \( MD \perp AC \).
(iii) Join MC. In \( \Delta AMD \) and \( \Delta CMD \):
\[ AD = CD \quad \text{(D is the mid-point of side AC)} \]
\[ \angle ADM = \angle CDM \quad \text{(Each } 90^{\circ} \text{)} \]
\[ DM = DM \quad \text{(Common)} \]
\[ \Rightarrow \Delta AMD \cong \Delta CMD \quad \text{(By SAS congruence rule)} \]
Therefore, \( AM = CM \) (By CPCT).
However, \( AM = \frac{1}{2}AB \) (M is the mid-point of AB).
Therefore, it can be said that \( CM = AM = \frac{1}{2}AB \).