## Quadrilaterals

**Chapter 8**

**Exercise 8.2**

**EX 8.2 QUESTION 1.**

**ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA (see Fig 8.29). AC is a diagonal. Show that:
(i) SR || AC and SR = 1/2 AC
(ii) PQ = SR
(iii) PQRS is a parallelogram.**

**Solution:**

(i) In ∆ACD,

∴ S is the mid-point of DA and

R is the mid-point of DC.

Hene, SR || AC and SR = ½AC …(1)[mid-point theorem]

(ii) In ∆ABC,

P is the mid-point of AB and Q is the mid-point of BC.

PQ = ½AC and PQ || AC …(2) [By mid-point theorem]

From (1) and (2), we get

PQ = ½AC = SR and PQ || AC || SR

⇒ PQ = SR and PQ || SR

(iii) In a quadrilateral PQRS,

PQ = SR and PQ || SR [Proved]

∴ PQRS is a parallelogram.

**EX 8.2 QUESTION 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.**

**Solution:**

**EX 8.2 QUESTION 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.**

**Solution:**

In ∆ABC, P is a midpoint of AB and Q is a midpoint of BC.

PQ = ½AC and PQ || AC …(1) [ mid-point theorem]

In ∆ACD, S is a midpoint of AD and R is a midpoint of CD.

SR =½AC and SR || AC …(2)

From (1) and (2),

PQ = SR and PQ || SR

∴ PQRS is a parallelogram.

Now, in ∆BCD, Q is a midpoint of BC and R is a midpoint of CD.

QR = ½AD …….(3) [mid-point theorem]

AC = BD………(4) [∵ Diagonals of a rectangle are equal]

From equation (1),(3) and (4)

⇒ PQ = QR

A parallelogram whose adjacent sides are equal is a rhombus. Hence, PQRS is a rhombus.

**EX 8.2 QUESTION 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 Fig. 8.30). Show that F is the mid-point of BC.**

**Solution:**

In ∆ABD, E is the mid-point of AD and EG || AB

G is the mid-point of BD.[converse of the mid-point theorem]

Again In ∆BCD, G is the midpoint of BD and FG || DC.

F is the mid-point of BC. [converse of the mid-point theorem]

**EX 8.2 QUESTION 5.**

**In a parallelogram ABCD, E and F are the mid-points of sides AB and CD respectively (see Fig. 8.31). Show that the line segments AF and EC trisect the diagonal BD.**

**Solution:**

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

∴ AB || DC

⇒ AE || FC …(1)

and AB = DC

⇒½ AB = ½ CD

⇒ AE = CF …(2)

From (1) and (2),

AE || PC and AE = PC

∴ ∆ECF is a parallelogram.

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

⇒ DP = PQ …(3) [By converse of mid-point theorem]

Similarly, In ∆BAP, E is the mid-point of AB and EQ || AP

⇒ 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.

**EX 8.2 QUESTION 6.**

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

**Solution:**

Let ABCD be a quadrilateral and P, Q, R and S are the midpoints of AB, BC, CD and DA respectively.

In ∆ABC, P is a mid-point AB and Q is a mid-point BC.

∴ PQ || AC and PQ = ½AC …(1) [mid-point theorem]

Similarly, RS || AC and RS = ½AC …(2)

From equation (1) and (2),

PQ || RS, PQ = RS

∴ PQRS is a parallelogram and diagonals of a parallelogram bisect each other

PR and SQ bisect each other.

**EX 8.2 QUESTION 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 = ½ AB**

**Solution:**

(i) In ∆ABC,

M is the mid-point of AB. [Given]

DM || BC , [Given]

D is the mid-point of AC. [Converse of mid-point theorem,]

(ii) ∠AMD = ∠ABC [Corresponding angles are equal] As

∠ADM = 90° [Given]

⇒ MD ⊥AC.

(iii) In ∆AMD and ∆CMD,

∠AMD = ∠CMD [Each equal to 90°]

DM = DM [Common]

∴ ∆ADM ≅ ∆CDM [SAS congruency]

⇒ AM = CM [C.P.C.T.] .. .(1)

AM = ½AB …(2)

From (1) and (2),

CM = AM = ½AB