Quadrilaterals

Quadrilaterals is a chapter in geometry that focuses on polygons with four sides. It builds upon the concepts of lines, angles, and triangles.

Key Topics

• Basic concepts: Definition of a quadrilateral, its properties, and different types (angle sum property, types of quadrilaterals).
• Properties of Parallelograms: Specific properties of parallelograms, including opposite sides, angles, and diagonals.
• Conditions for a Quadrilateral to be a Parallelogram: Various conditions that guarantee a quadrilateral to be a parallelogram.
• Rectangles, Rhombuses, and Squares: Properties and relationships between these special types of parallelograms.
• Kite: Definition and properties of a kite.

Core Ideas

• Identifying the necessary conditions for a quadrilateral to be a parallelogram.
• Applying the properties of quadrilaterals to solve geometric problems.

This chapter provides a foundation for understanding more complex geometric shapes and their properties.

Exercise 8.1

1. If the diagonals of a parallelogram are equal, then show that it is a rectangle.

Ans : 

In ∆ABC and ∆DCB,

AC = DB [Given]

AB = DC [Opposite sides of a parallelogram]

BC = CB [Common]

∴ ∆ABC ≅ ∆DCB [By SSS congruency]

⇒ ∠ABC = ∠DCB [By C.P.C.T.] …(1)

Now, AB || DC and BC is a transversal. [ ∵ ABCD is a parallelogram]

∴ ∠ABC + ∠DCB = 180° … (2) [Co-interior angles]

From (1) and (2), we have

∠ABC = ∠DCB = 90°

∴ ABCD is a rectangle.

2. Show that if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus.

Ans : 

∴ In ∆AOB and ∆AOD, we have

AO = AO [Common]

∠AOB = ∠AOD [Each 90]

∴ ∆AQB ≅ ∆AOD [By,SAS congruency

∴ AB = AD [By C.P.C.T.] ……..(1)

Similarly, AB = BC .. .(2)

BC = CD …..(3)

CD = DA ……(4)

AB = BC = CD = DA

Thus, the quadrilateral ABCD is a rhombus.

3. Diagonal AC of a parallelogram ABCD bisects ∠A (see figure). Show that

(i) it bisects ∠C also,

(ii) ABCD is a rhombus.

Ans : 

(i) AC bisects angle C

• Since ABCD is a parallelogram, AB is parallel to DC and AD is parallel to BC.
• Therefore, angle DAC is equal to angle BCA (alternate interior angles).
• Similarly, angle BAC is equal to angle DCA (alternate interior angles).
• But angle DAC is equal to angle BAC (given).
• Hence, angle BCA is equal to angle DCA.

Therefore, AC bisects angle C.

(ii) ABCD is a rhombus

• We know that AC bisects both angle A and angle C.
• This means triangle ABC is congruent to triangle ADC (ASA congruence).
• Therefore, AB = CD and BC = AD (corresponding parts of congruent triangles).
• Since ABCD is a parallelogram, 
• AB = CD and AD = BC.
• Combining the above equalities, we get AB = BC = CD = AD.

Therefore, ABCD is a rhombus.

4. ABCD is a rectangle in which diagonal AC bisects ∠A as well as ∠C. Show that

(i) ABCD is a square

(ii) diagonal BD bisects ∠B as well as ∠D.

Ans : 

Proof:

i) ABCD is a square

• Since AC bisects angle A, angle BAC = angle DAC.
• Since ABCD is a rectangle, 
• angle A and angle C are right angles.
• Therefore, angle BAC = angle DAC = 45°.
• Similarly, angle BCA = angle DCA = 45°.
• In triangle ABC, angle ABC = 180° – (angle BAC + angle BCA) = 180° – 90° = 90°.
• So, all angles of ABCD are 90°, and it is a rectangle with all sides equal.
• Hence, ABCD is a square.

ii) 

Since ABCD is a square, all its sides are equal and all its angles are right angles.

• Consider triangles ABD and CBD.
• AB = BC (sides of a square)
• AD = CD (sides of a square)
• BD is common to both triangles.
• Therefore, by SSS congruence, triangle ABD is congruent to triangle CBD.
• Hence, angle ABD = angle CBD and angle ADB = angle CDB.

Therefore, diagonal BD bisects both angle B and angle D.

5. In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP = BQ (see figure). Show that

Ans : 

Proof:

(i) ΔAPD ≅ ΔCQB

• Given: ABCD is a parallelogram, DP = BQ.
• Proof:
  • In triangles APD and CQB,
    • AD = BC (opposite sides of a parallelogram)
    • ∠ADP = ∠CBQ (alternate interior angles as AD || BC and BD is the transversal)
    • DP = BQ (given)
  • Therefore, by SAS congruence, ΔAPD ≅ ΔCQB.

(ii) AP = CQ

• Since ΔAPD ≅ ΔCQB (proved in (i)), their corresponding parts are equal.
• Therefore, AP = CQ.

(iii) ΔAQB ≅ ΔCPD

• Proof:
  • In triangles AQB and CPD,
    • AB = CD 
    • ∠ABQ = ∠CDP (alternate interior angles as AB || CD and BD is the transversal)
    • BQ = DP (given)
  • Therefore, by SAS congruence, ΔAQB ≅ ΔCPD.

(iv) AQ = CP

• Since ΔAQB ≅ ΔCPD (proved in (iii)), their corresponding parts are equal.
• Therefore, AQ = CP.

(v) APCQ is a parallelogram

• From (ii) and (iv), we have AP = CQ and AQ = CP.
• Therefore, APCQ is a parallelogram.

6. ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD (see figure). Show that

Ans : 

Proof:

(i) In △APB and △CQD,

∠APB = ∠CQD = 90° (Each 90∘)

AB = CD 

∠ABP = ∠CDQ (Alternate interior angles for AB∥CD)

∴ △APB ≅ △CQD (By AAS congruency)   

(ii) AP = CQ (By CPCT)

Therefore, AP = CQ.

7. ABCD is a trapezium in which AB || CD and AD = BC (see figure). Show that

(i )∠A=∠B

(ii )∠C=∠D

(iii) ∆ABC ≅ ∆BAD

(iv) diagonal AC = diagonal BD

Ans : 

(i) ∠A = ∠B

• Since AB is parallel to CD, AD is a transversal.
• Therefore, ∠A and ∠B are alternate interior angles.
• Hence, ∠A = ∠B.

(ii) ∠C = ∠D

• Similarly, since AB is parallel to CD, BC is a transversal.
• Therefore, ∠C and ∠D are alternate interior angles.
• Hence, ∠C = ∠D.

(iii) ΔABC ≅ ΔBAD

• In triangles ABC and BAD,
  • AB is common.
  • BC = AD (given)
  • ∠A = ∠B (proved in (i))
• Therefore, by the SAS congruence rule, ΔABC ≅ ΔBAD.

(iv) Diagonal AC = diagonal BD

• Since ΔABC ≅ ΔBAD, their corresponding parts are equal.
• Therefore, AC = BD.

Exercise 8.2

1. ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA (see figure). AC is a diagonal. Show that

(i) SR || AC and SR = 1/2 AC

(ii) PQ = SR

(iii) PQRS is a parallelogram.

Ans : 

Proof:

(i) Consider triangle ADC.

• By the mid-point theorem, SR || AC and SR = 1/2 AC.

(ii) PQ = SR

• Similarly, consider triangle ABC.
• By the mid-point theorem, PQ || AC and PQ = 1/2 AC.
• From (i) and (ii), we have SR || AC and PQ || AC.
• Therefore, SR || PQ.
• Also, SR = 1/2 AC and PQ = 1/2 AC, so SR = PQ.

(iii) PQRS is a parallelogram

• From (ii), we have PQ || SR and PQ = SR.
• Therefore, PQRS is a parallelogram.

2. 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.

Ans : 

1. PQRS is a Parallelogram:

Using the mid-point theorem, we can show that PQ is parallel to SR and PQ = SR. 

Similarly, PS is parallel to QR 

and PS = QR.

Since one pair of opposite sides is parallel and equal, PQRS is a parallelogram.

2. Diagonals of PQRS are equal:

In rhombus ABCD, AC = BD (diagonals of a rhombus are equal).

Since P, Q, R, and S are midpoints, PR = 1/2 AC and QS = 1/2 BD.

Therefore, PR = QS.

3. Conclusion:

A parallelogram with equal diagonals is a rectangle.

Since PQRS is a parallelogram with equal diagonals PR and QS, it is a rectangle.

Hence, PQRS is a rectangle.

3. ABCD is a rectangle and P, Q, R ans S are mid-points of the sides AB, BC, CD and DA, respectively. Show that the quadrilateral PQRS is a rhombus.

Ans :

1. PQRS is a Parallelogram:

• Using the mid-point theorem, we can show that PQ is parallel to SR and PQ = SR. 
• Similarly, PS is parallel to QR 
• and PS = QR.
• Since one pair of opposite sides is parallel and equal, PQRS is a parallelogram.

2. Diagonals of PQRS are equal and perpendicular:

• In rectangle ABCD, AC = BD (diagonals of a rectangle are equal).
• Since P, Q, R, and S are midpoints, PR = 1/2 AC and QS = 1/2 BD.
• Therefore, PR = QS.
• Also, in a rectangle, the diagonals intersect at right angles. Therefore, PR and QS are perpendicular to each other.

3. Conclusion:

• A parallelogram with equal and perpendicular diagonals is a rhombus.
• Since PQRS is a parallelogram with equal and perpendicular diagonals PR and QS, it is a rhombus.

Hence, PQRS is a rhombus.

4. ABCD is a trapezium in which AB || 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 figure). Show that F is the mid-point of BC.

Ans : 

Step 1: Introduction of Point G

Let the line EF intersect BD at point G.

Step 2: Applying the Midpoint Theorem

• In triangle ABD, E is the midpoint of AD and EG is parallel to AB. Therefore, by the converse of the mid-point theorem, G is the midpoint of BD.

Step 3: Applying the Midpoint Theorem Again

• In triangle BCD, GF is parallel to CD (as EF is parallel to AB, and AB is parallel to CD) and G is the midpoint of BD. Therefore, by the converse of the mid-point theorem, F is the midpoint of BC.

Conclusion:

Hence, F is the midpoint of BC.

5. In a parallelogram ABCD, E and F are the mid-points of sides AB and CD respectively (see figure). Show that the line segments AF and EC trisect the diagonal BD.

Ans : 

Since, the opposite sides of a parallelogram are parallel and equal.

∴ AB || DC

⇒ AE || FC …(1)

and AB = DC

⇒ 1/2AB = 1/2DC

⇒ AE = FC …(2)

From (1) and (2), we have

AE || PC and AE = PC

∴ ∆ECF is a parallelogram.

Now, in ∆DQC, we have F is the mid-point of DC and FP || CQ

[∵ AF || CE]

⇒ DP = PQ …(3)

⇒ BQ = PQ …(4)

[By converse of mid-point theorem]

∴ From (3) and (4), we have

DP = PQ = BQ

So, the line segments AF and EC trisect the diagonal BD.

6. 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 ⊥ AC

(iii) CM = MA = 1/2AB

Ans : 

Proof:

i) D is the mid-point of AC

• Since MD is parallel to BC and M is the midpoint of AB, by the converse of the mid-point theorem, D is the midpoint of AC.

ii) MD ⊥ AC

• Since MD is parallel to BC and AC is a transversal, the corresponding angles ∠MDC and ∠ACB are equal.
• As ∠ACB is a right angle (given), ∠MDC is also a right angle.
• Therefore, MD is perpendicular to AC.

iii) CM = MA = 1/2 AB

• Since M is the midpoint of AB, MA = MB = 1/2 AB.
• From part (i), 
• In triangles AMD and CMD,
  • AM = CM (proved above)
  • AD = CD (proved in part (i))
  • ∠AMD = ∠CMD (both right angles)
• Therefore, by the SAS congruence rule, triangle AMD is congruent to triangle CMD.
• Hence, MD = CM.
• Since MA = MB and CM = MD, we have CM = MA = 1/2 AB.