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**Question 1. ABCD is a quadrilateral in which P, Q, R and S are midpoints of the sides AB, BC, CD and DA (see figure). AC is a diagonal. Show that: (i) SR || AC and SR = **(1/2)

(ii) PQ = SR (iii) PQRS is a parallelogram.

**Solution:** We have P as the mid-point of AB, Q as the mid-point of BC, R as the mid-point of CD, S as the mid-point of DA, and AC as the diagonal of quadrilateral ABCD.

(i) To prove that SR =(1/2) AC and SR || AC.

In Î”ACD, we have

S as the mid-point of AD, R as the mid-point of CD.

âˆµ The line segment joining the mid-point of any two sides of a triangle is parallel to the third side and half of it.

âˆ´ SR = (1/2)AC and SR || AC

(ii) To prove that PQ = SR.

In Î”ABC, we have P is the mid-point of AB, Q is the mid-point of BC.

âˆ´ PQ = (1/2) AC â€¦(1)

Also, SR = (1/2) AC [Proved] â€¦(2)

From (1) and (2), PQ = SR

(iii) To prove that PQRS is a parallelogram.

In Î”ABC, P and Q are the mid-points of AB and BC.

âˆ´ PQ = (1/2)AC and PQ || AC ...(3)

In Î”ACD, S and R are the mid-points of DA and CD.

âˆ´ SR = (1/2)AC and SR || AC â€¦(4)

From (3) and (4), we get

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

â‡’ PQ = SR and PQ || SR

i.e. One pair of opposite sides in quadrilateral PQRS is equal and parallel.

âˆ´ 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. Solution: **We have P as the mid-point of AB, Q as the midpoint of BC, R as the mid-point of CD, S as the mid-point of DS.

We have to prove that PQRS is a rectangle.

Let us join AC.

âˆµ In Î”ABC, P and Q are the mid-points of AB and BC.

âˆ´ PQ =(1/2)AC and PQ || AC â€¦(1)

Also in Î”ADC, R and S are the mid-points of CD and DA.

âˆ´ SR = (1/2)AC and SR || AC

From (1) and (2), we get

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

â‡’ PQ = SR and PQ || SR

i.e. One pair of opposite sides of quadrilateral PQRS is equal and parallel.

âˆ´ PQRS is a parallelogram.

Now, in Î”ERC and Î”EQC, âˆ 1= âˆ 2 [âˆµ The diagonal of a rhombus bisects the opposite angles]

CR = CQ [Each is equal to(1/2) of a side of rhombus]CE = CE [Common]

âˆ´ Î”ERC â‰Œ Î”EQC [SAS criteria]

â‡’ âˆ 3= âˆ 4[c.p.c.t.]

But âˆ 3 + âˆ 4 = 180Âº [Linear pair] â‡’ âˆ 3= âˆ 4 = 90Â°

But âˆ 5= âˆ 3 [Vertically opposite angles]

âˆ´ âˆ 5 = 90Âº PQ || AC

â‡’ PQ || EF

âˆ´ PQEF is a quadrilateral having a pair of opposite sides parallel and one of the angles is 90Âº.

âˆ´ PQEF is a rectangle.

â‡’ âˆ RQP = 90Âº

âˆ´ One angle of parallelogram PQRS is 90Âº.

Thus, PQRS is a rectangle.

**Question 3. ABCD is a rectangle 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 rhombus. Solution: **In a rectangle ABCD, P is the mid-point of AB, Q is the midpoint of BC, R is the mid-point of CD, S is the mid-point of DA AC is the diagonal.

Now, in Î”ABC,

PQ =(1/2)AC and PQ || AC

[Mid-point theorem] â€¦(1)

Similarly, in Î”ACD,

SR =(1/2)AC and SR || AC â€¦(2)

From (1) and (2), we get PQ = SR and PQ || SR Similarly, by joining BD,

we have PS = QR and PS || QR

i.e. Both pairs of opposite sides of quadrilateral PQRS are equal and parallel.

âˆ´ PQRS is a parallelogram.

Now, in Î”PAS and Î”PBQ, âˆ A= âˆ B [Each = 90Âº] AP = BP [Each = (1/2)AB]

AS = BQ [Each =(1/2) of opposite sides of a rectangle]

âˆ´ Î”PAS â‰Œ Î”PBQ [SAS criteria]

âˆ´ Their corresponding parts are equal.

â‡’ PS = PQ Also PS = QR [Proved] and

PQ = SR [Proved]

âˆ´ PQ = QR = RS = SP

i.e. PQRS is a parallelogram having all of its sides equal.

â‡’ 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 figure). Show that F is the mid-point of BC.**

**Solution:** In trapezium ABCD, AB || DC. E is the mid-point of AD.

EF is drawn parallel to AB. We have to prove that F is the mid-point of BC.

Join BD.

In DDAB,

âˆµ E is the mid-point of AD [Given]

and EG || AB [âˆµ EF || AB]

âˆ´ Using the converse of mid-point theorem, we get that G is the mid-point BD.

Again in DBDC,

âˆµ G is the mid-point of BD [Proved]

GF || DC [âˆµ AB || DC and EF || AB and GF is a part of EF]

âˆ´ Using the converse of the mid-point theorem, we get that F is the mid-point of BC.

**Question 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. Solution:** We have ABCD is a parallelogram such that: E is the mid-point of AB and F is the mid-point of CD.

Let us join the opposite vertices B and D.

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

âˆ´ AB || DC â‡’ AE || FC â€¦(1)

Also AB = DC

or (1/2) AB =(1/2)DC â‡’ AE = FC â€¦(2)

From (1) and (2),

we can say that AECF is quadrilateral having a pair of the opposite sides as parallel and equal.

âˆ´ AEFC is a parallelogram.

â‡’ AE || CF

Now, in DDBC, F is the mid-point of DC [Given]

and FP || CQ [âˆµ AF || CE]

â‡’ P is the mid-point of DQ [Converse of mid-point theorem]

â‡’ DP = PQ â€¦(3)

Similarly, in DBAP, BQ = PQ â€¦(4)

âˆ´ From (3) and (4), we have

DP = PQ = BQ

â‡’ The line segments AF and EC trisect the diagonal BD.

**Example 6. Show that the line segments joining the mid-points of the opposite sides of a quadrilateral bisect each other. Solution: **A quadrilateral ABCD such that the mid-points of AB, BC, CD and DA are P, Q, R and S respectively, we have to prove that diagonals of PQRS are bisected at O.

Join PQ, QR, RS and SP. Let us also join PR and SQ.

Now, in Î”ABC, we have P and Q as the mid-points of its sides AB and BC respectively.

âˆ´ PQ || AC and PQ =(1/2)AC

Similarly, RS || AC and RS =(1/2)AC

â‡’ PQRS is a quadrilateral having a pair of opposite sides (PQ and RS) as equal and parallel.

âˆ´ PQRS is a parallelogram.

But the diagonals of a parallelogram bisect each other. i.e. PR and SQ bisect each other.

Thus, the line segments joining the mid-points of opposite sides of a quadrilateral ABCD bisect each other.

**Question 7. ABC is a triangle, right angled at C. A line through the midpoint M of hypotenuse AB and parallel to BC intersects AC at D. Show that (i) D is the mid-point of AC (ii) MD âŠ¥ AC (iii) CM = MA =**(1/2)** AB Solution:** We have a triangle ABC, such that âˆ C = 90Âº M is the mid-point of AB and MD || BC

(i) To prove that D is the mid-point of AC.

In Î”ACB, we have M as the mid-point of AB. [Given]

MD || BC [Given]

âˆ´ Using the converse of mid-point theorem, D is the mid-point of AC.

(ii) To prove that MD âŠ¥ AC.

Since, MD || BC [Given] and AC is a transversal.

âˆ´ âˆ MDA = âˆ BCA [Corresponding angles]

But âˆ BCA = 90Âº [Given]

âˆ´ âˆ MDA = 90Âº â‡’ MD âŠ¥ AC.

(iii) To prove that CM = MA =(1/2) AB

In Î”ADM and Î”CDM, we have

âˆ ADM = âˆ CDM [Each = 90Âº]

MD = MD [Common]

AD = CD [âˆµ M is the mid-point of AC (Proved)]

âˆ´ Î”ADM â‰Œ Î”COM [SAS criteria]

â‡’ MA = MC [c.p.c.t.] â€¦(1)

âˆµ M is the mid-point AB. [Given]

âˆ´ MA =(1/2)AB â€¦(2)

From (1) and (2), we have

CM = MA = (1/2) AB

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