## Chapter 8 Quadrilaterals Exercise 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

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

ii) PQ = SR

iii) PQRS is a parallelogram

**Answer:**

i) In ∆ADC, S and R are the midpoints of AD and CD respectively.

Therefore, by midpoint theorem (the line segment joining the midpoints of two sides of a triangle is parallel to the third side and also half of it).

∴ SR || AC and SR = 1/2 AC …(1)

ii) In ∆ABC,

P and Q are midpoints of AB and BC respectively.

Therefore, by midpoint theorem,

∴ PQ || AC and PQ = 1/2 AC …(2)

From (1) and (2),

PQ || SR and PQ = SR …(3)

iii) From (3), PQ || SR and PQ = SR. As one pair of opposite sides of quadrilateral PQRS is equal and parallel.

Hence PQRS is a parallelogram.

**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.**

Answer:

Answer:

In ∆ABC, P and Q are the midpoint of AB and BC respectively.

→ PQ || AC and PQ = 1/2 AC (by midpoint theorem) …(1)

In ∆ADC, R and S are the midpoint of CD and AD respectively.

→ RS || AC and RS = 1/2 AC (by midpoint theorem) …(2)

From (1) and (2),

PQ || 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 O.

In quadrilateral OMQN,

MQ || ON (∵ PQ || AC)

QN || OM (∵ QR || BD)

∴ OMQN is a parallelogram

→ ∠MQN = ∠NOM

→ ∠PQR = ∠NOM

However, ∠NOM = 90° (diagonals of a rhombus are perpendicular to each other)

Hence ∠PQR = 90°

PQRS is a parallelogram with one of its interior angles as 90°. Hence, PQRS is a rectangle.

**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.**

Answer:

Answer:

Let us join BD and AC.

In ∆ABC, P and Q are midpoints of AB and BC respectively.

→ PQ || AC and PQ = 1/2 AC (by mid point theorem) …(1)

Similarly, in ∆ADC, S and R are midpoints of AD and DC respectively.

→ SR || AC and SR = 1/2 AC (by mid point theorem) …(2)

From (1) and (2), PQ || 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 || QR and PS = QR (opposite sides of parallelogram) …(3)

In ∆BCD, Q and R are midpoints of BC and CD respectively.

→ QR || BD and QR = 1/2 BD (by midpoint theorem) …(4)

However, the diagonals of a rectangle are equal.

→ AC = BD …(5)

From (1), (2), (3), (4) and (5), we get

PQRS is a rhombus.

**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.**

**Answer:**By converse of midpoint theorem, we know that a line drawn through the midpoint of any side of a triangle and parallel to another side bisects the third side.

In ∆ABD, EF || AB and E is the midpoint of AD.

∴ G will be midpoint of DB (as EF || AB and AB || CD).

Hence, EF || CD (two lines are parallel to each other)

In ∆BCD, GF || CD and G is midpoint of BD.

∴ By converse of midpoint theorem, F is midpoint of BC.

**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.**

**Answer:**

→ AB || CD

Hence, AE || FC

Again AB = CD (opposite sides of parallelogram ABCD)

= 1/2 AB = 1/2 CD

= AE = FC (as E & F are midpoints of AB and CD respectively)

In quadrilateral AECF, one pair of opposite sides (AE and FC) are parallel and equal to each other.

∴ AECF is a parallelogram.

= AF || EC (opposite sides of parallelogram AECF)

In ∆DQC, F is midpoint of DC and FP || CQ (as AF || EC)

∴ By converse of midpoint theorem, it can be said that P is the midpoint of DQ

Therefore, DP = PQ …(1)

In ∆APB, E is the midpoint of AB and EQ || AP (as AF || EC)

Therefore by converse of midpoint theorem, it can be said that Q is the midpoint of PB.

Therefore, PQ = QB …(2)

From (1) and (2),

DP = PQ = BQ

Hence the line segment AF and EC trisect the diagonal BD.

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

Answer:Let ABCD be a quadrilateral in which P, Q, R, S are midpoints of AB, BC, CD and DA respectively. Join PQ, QR, RS, SP and BD.

Answer:

In ∆ABD, S and P are midpoints of AD and AB respectively.

Therefore, by midpoint theorem,

SP || BD and SP = 1/2 BD …(1)

In ∆BCD, Q and R are midpoints of BC and CD respectively.

Therefore, by midpoint theorem,

QR || BD and QR = 1/2 BD …(2)

From (1) and (2)

SP || QR and SP = QR

In quadrilateral SPQR, one pair of opposite sides is equal and parallel to each other.

∴ SPQR is a parallelogram.

We know that the diagonals of a parallelogram bisect each other.

Therefore, PR and QS bisect each other.

**i) It is given that in ∆ABC, M is the midpoint of AB and MD || BC.**

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 = 1/2 AB

Answer:

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 = 1/2 AB

Answer:

Therefore, D is the midpoint of AC (by converse of midpoint theorem)

ii) As DM || CB and AC is a transversal

→ ∠MDC + ∠DCB = 180° (co-interior angles)

= ∠MDC + 90° = 180°

= ∠MDC = 90°

→ MD ⊥ AC

iii) Join CM.

In ∆AMD and ∆CMD,

AD = AD (D is midpoint of AC)

∠ADM = ∠CDM (each 90°)

DM = DM (common)

Therefore, by SAS congruency, ∆AMD ≅ ∆CMD.

By CPCT, AM = CM

However, AM = 1/2 AB (as M is the midpoint of AB)

Therefore, it can be said that

CM = AM = 1/2 AB

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