# Quadrilaterals – Exercise 8.2 – (MATHEMATICS) – 9th Class

**Exercise 8.2**

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

(i) and SR =

(ii) PQ = SR

(iii) PQRS is a parallelogram.

### Show Answer

**Sol.** (i) Consider triangle ACD,

S and R are mid-points of sides AD and DC respectively.

∴ and SR = ….(i)

[Line segment joining mid-points of two sides of a triangle is parallel to the third and half of it.]

(ii) Consider triangle ABC, P and Q are mid-points of sides AB and BC respectively.

∴ and PQ= ….(ii)

[Reason same as above]

From (i) and (ii),

and ⇒ …(iii)

and SR = ** **AC and PQ = AC ⇒ SR = PQ. …(iv)

(iii) and SR = PQ. [From (iii) and (iv)]

⇒ PQRS is a parallelogram.

**2. ABCD is a rhombus and P, Q, R and S are the midpoints of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rectangle.**

### Show Answer

**Sol.**First prove that PQRS is a parallelogram.

(i) Consider triangle ACD,

S and R are mid-points of sides AD and DC respectively.

∴ and

[Line segment joining mid-points of two sides of a triangle is parallel to the third and half of it.]

(ii) Consider triangle ABC, P and Q are mid-points of sides AB and BC respectively.

∴ and …(ii) [Reason same as above]

From (i) and (ii),

and ⇒ …(iii)

and and ⇒ SR = PQ. …(iv)

(iii) and SR = PQ. [From (iii) and (iv)]

⇒ PQRS is a parallelogram.

PQRS is a parallelogram.

As and is a parallelogram.

⇒ ∠YPX = ∠YOX = 90° [Diagonals of a rhombus bisect each other and are at right angles.]|

As in parallelogram PQRS,

∠SPQ is 90°.

∴ PQRS is a rectangle.

**3. 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.**

**Show Answer**

**Join AC and BD.**

Sol. Construction:

Sol. Construction:

As ABCD is a rectangle.

∴ AC = BD …(i)

Consider ΔABC, P and Q are midpoints of sides AB and BC respectively.

∴ and

Similarly, consider AADC, S and R are mid-points of sides AD and DC respectively.

∴ and ….(iii)

From (ii) and (iii)

……(iv)

Similarly, we can show

From (i), (iv) and (v), we have PQ = QR .= RS = SP

∴ PQRS is a rhombus.

**4. ABCD is a trapezium in which , 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.**

### Show Answer

**Sol.**Consider ΔADB, ⇒

⇒ G is mid-point of BD. …(i)

[ A line drawn through mid-point of one side, parallel to other bisects the third side.]

Consider triangle BCD,

and

⇒ ⇒

⇒ F is mid-point of BC. [Reason same as above]

**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.**

### Show Answer

**Sol.**

⇒ AE = CF

As AE = CF and [ ]

⇒ AECF is a parallelogram.

⇒ …(i)

Consider triangle ABP,

E is mid-point of AB and [From (i)]

⇒ Q is mid-point of BP [A line segment drawn through mid-point of one side of a triangle and parallel to other, bisects the third side.]

BQ = PQ …(ii)

Similarly, by considering triangle DCQ and proceeding as

above, we can show that

DP = PQ …(iii)

⇒ BQ = PQ = DP [From (ii) and (iii)]

⇒ P and Q trisect BD.

**6. Show that the line segments joining the mid-points of the opposite sides of a quadrilateral bisect each other.**

### Show Answer

**Sol.**

(i) Consider triangle ACD,

S and R are mid-points of sides AD and DC respectively.

∴ and SR …(i)

[Line segment joining mid-points of two sides of a triangle is parallel to the third and half of it.]

(ii) Consider triangle ABC, P and Q are mid-points of sides AB and BC respectively.

∴ and PQ [Reason same as above]

From (i) and (ii),

and ⇒ …(iii)

and and ⇒ SR = PQ . …(iv)

(iii) and SR = PQ. [From (iii) and (iv)]

⇒ PQRS is a parallelogram.

We know that diagonals of a parallelogram bisect each other, i.e., OP = OR and OQ = OS.

Hence, line segments joining midpoints of opposite sides of a quadrilateral bisect each other.

**7. 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 = AB.

### Show Answer

**Sol.**

(i) meets AC at D.

∴ D is mid-point of AC.

[A line through the mid-point of a ID side of a triangle parallel to other bisects and third side.]

(ii) and AC is transversal.

∴ ∠ADM = ∠ACB. [ Corresponding angles ]

⇒ ∠ADM = 90° [ ∠ACB = 90° ]

⇒ MD ⊥ AC.

(iii) Consider triangles ADM and CDM,

AD = DC [ From result (i) ]

MD is common.

∠ADM = ∠CDM [90° each] [From result (ii)]

∴ ΔADM = ΔCDM [SAS]

∴ MA = CM = [ M is mid-point of AB]