**Q.1: In the adjoining figure ABCD is a trapezium in which AB∥DC and E is the midpoint of AD. A line segment EF∥AB meets BC at F. Show that F is the midpoint of BC.**

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**Sol:**** **

Join BD to cut EF at M

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Now, in

M is the midpoint of BD. (By converse of midpoint theorem)

Again, in

Therefore, F is the midpoint of BC (By converse of midpoint theorem)

**Q.2: In the adjoining figure ABCD is a parallelogram in which E and F are the midpoints of AB and CD respectively. If GH is a line segment that cuts AD, EF and BC at G, P and H respectively, prove that GP = PH.**

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**Sol:**

In parallelogram ABCD, we have:

AD || BC and AB || DC

AD = BC and AB = DC

AB = AE + BE and DC = DF + FC

AE = BE = DF = FC

Now, DF = AE and OF || AE.

i.e., AEFD is a parallelogram.

AD || EF

Similarly, BEFC is also a parallelogram.

EF || BC and AD || EF || BC

Thus, AD, EF and BC are three parallel lines cut by the transversal line DC at D, F and C, respectively such that DF = FC.

These lines AD, EF and BC are also cut by the transversal AB at A, E and B, respectively such that AE = BE

Similarly, they are also cut by GH

Therefore, GP = PH (By intercept theorem)

**Q.3: In the adjoining figure ABCD is a trapezium in which AB∥DC and P and Q are the midpoints of AD and BC respectively. DQ and AB when produced meet at E. Also AC and PQ intersect at R . Prove that **

**(i) DQ = QE**

**(ii) PR∥AB**

**(iii) AR∥RC**

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**Sol:**

Given: AB || DC, AP = PD and BQ = CQ

**(i)** In

∠DQC = ∠BQE (Vertically opposite angles)

∠DCQ = ∠EBQ (Alternate angles, as AE || DC)

BQ = CQ (P is the midpoints)

Hence, DQ = QE (CPCT)

**(ii)** Now, in

PQ || AE, PQ AB || DC and AB || PR || DC

**(iii)** PQ AB and DC are the three lines cut by transversal AD at P such that AP = PD. These lines PQ, AB, DC are also cut by transversal BC at Q such that BQ = QC. Similarly, lines PQ, AB and DC are also cut by AC at R. Therefore, AR = RC (By intercept theorem).

**Q.4: In the adjoining figure AD is a median of △ABC and DE∥BA. Show that BE is also a median of triangle △ABC **

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**Sol:**

AD is a median of

We know that the line drawn through the midpoint of one side of a triangle and parallel to another side bisects the third side.

Here, in

i.e., AE = EC

E is the midpoint of AC

Therefore, BE is the median of

**Q.5: In the adjoining figure AD and BE are the media****ns of △ABC and DF∥BE. Show that CF=14AC.**

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**Sol:**

In

AC = AE + EC . . . . . . (i)

AE = EC . . . . . . (ii) [BE is the median of ABDC]

AC = 2EC . . . . . . . (iii)

In

EF = CF (By midpoint theorem, as D is the midpoint of BC)

But EC = EF + CF

EC = 2 (CF) . . . . . . (iv)

From (iii) and (iv), we get:

AC = 2 (2CF)

Therefore, CF = 14 AC

**Q.6: In the adjoining figure ABCD is a parallelogram, E is the midpoint of DC and through D, a line segment is drawn parallel to EB to meet CB produced at G and it cuts AB at F. Prove that:**

**(i). q AB=12GC**

**(ii). DG = 2 EB**

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**Sol:**

(i) In

DG || EB and DE = EC (E is the midpoint of DC)

Also, GB = BC (By midpoint theorem)

B is the midpoint of GC

Now, GC = GB + BC

GC = 2BC = 2 (AD)

AD = 12 GC

In ADCG, DG || EB and E is the midpoint of DC and B is the midpoint of GC.

By midpoint theorem, the line segment joining the mid points of any two sides of a triangle is parallel to the third side and is half of it.

i.e., EB = 12DG

DG = 2 (EB)

**Q.7: Prove tha****t the line segment joining the midpoints of the sides of a triangle divided it into 4 congruent triangles.**

**Sol:**

D, E and F are the midpoints of sides AB, BC and CA, respectively

As, D and E are the mid points of sides AB, and BC of

DE || AC (By midpoint theorem)

Similarly, DF || BC and EF || AB

Therefore, ADEF, BDFE, and DFCE are all parallelograms.

Now, DE is the diagonal of the parallelogram BDFE.

Similarly, DF is the diagonal of the parallelogram ADEF.

And, EF is the diagonal of the parallelogram DFCE.

So, all the four triangles are congruent.

**Q.8: In the adjoining figure D,E and F are the midpoints of the sides BC CA and AB respectively of △ABC. Show that ∠EDF=∠A,∠DEF=∠Band∠DFE=∠C.**

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**Sol:**

As F and E are the mid points of sides AB and AC of

ABC.

FE || BC (By mid point theorem)

Similarly, DE || FB and FD || AC

Therefore, AFDE, BDEF, and DCEF are all parallelograms.

In parallelogram AFDE, we have:

∠A = ∠EDF (Opposite angles are equal)

In parallelogram BDEF, we have:

∠B = ∠DEF (Opposite angles are equal)

In parallelogram DCEF, we have:

∠C = ∠DFE (Opposite angles are equal)

**Q.9: Show that the quadrilateral formed by joining the midpoints of the pairs of adjacent sides of a rectangle is a rhombus.**

**Sol:**

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

Join AC, a diagonal of the rectangle.

In

PQ || AC and PQ = 12AC [By midpoint theorem]

Again, in

SR || AC and SR = 12AC [By midpoint theorem]

Now, PQ || AC and SR || AC

PQ || SR

Also, PQ = SR [Each equal to 12 AC] . . . . . . . (i)

So, PQRS is a parallelogram.

Now, in

AS = BQ, ∠A = ∠B = 90° and AP = BP

i.e.,

PS = PQ . . . . . . . . . (ii)

Similarly, ASDR = ACICR

SR = RQ . . . . . . . . (iii)

**From (i), (ii) and (iii), we have: **

PQ = PQ = SR = RQ

Hence, PQRS is a rhombus.

**Q.10: Show that the quadrilateral formed by joining the midpoints of the pairs of adjacent sides of a rhombus is a rectangle.**

**Sol:**

Let, ABCD be a rhombus and P, Q, R and S be the midpoints of AB, BC, CD and DA, respectively. Join the diagonals, AC and BD.

In

PQ || AC and PQ = 12AC [By midpoint theorem]

Again, in ADAC, the points S and R are the midpoints of AD and DC, respectively.

SR || AC and SR = 12AC [By midpoint theorem]

Now, PQ || AC and SR || AC => PQ || SR

Also, PQ = SR [Each equal to 12 AC] . . . . . . . . . (i)

So, PQRS is a parallelogram.

We know that the diagonals of a rhombus bisect each other at right angles ∠EOF = 90°.

Now, RQ || DB => RE || FO

Also, SR || AC => FR || OE

Therefore, OERF is a parallelogram.

So, ∠FRE = ∠EOF = 90 (Opposite angles are equal)

Thus, PQRS is a parallelogram with ∠R = 90°.

Therefore, PQRS is a rectangle.

**Q.11: Show that the quadrilateral formed by joining the midpoints of the pairs of adjacent sides of a square is a square **

**Sol:**

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

Join the diagonals AC and BD. Let BD cut SR at F and AC cut RQ at E. Let 0 be the intersection point of AC and BD.

In

PQ || AC and PQ = 12AC [By midpoint theorem]

Again, in ADAC, the points S and R are the midpoints of AD and DC, respectively. SR || AC and SR = 12AC [By midpoint theorem]

Now, PQ || AC and SR || AC => PQ || SR

Also, PQ = SR [Each equal to 12 AC]

So, PQRS is a parallelogram.

Now, in

AS = BQ

AP = BP

i.e.,

PS = PQ . . . . . . . . (ii)

Similarly,

SR = RQ . . . . . . . . . . (iii)

From (i), (ii) and (iii), we have:

PQ = PS = SR = RQ . . . . . . . . (iv)

We know that the diagonals of a square bisect each other at right angles.

∠EOF = 90°

Now, RQ || DB

=> RE || FO

Also, SR || AC

=>FR || OE

Therefore, OERF is a parallelogram.

So, ∠FRE = ∠EOF = 90° (Opposite angles are equal)

Thus, PQRS is a parallelogram with ∠R = 90° and PQ = PS = SR = RQ.

Therefore, PQRS is a square.

**Q.12: Prove that the line segment joining the midpoints of opposite sides of a quadrilateral bisect each other.**

**Sol:**

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

Join PQ, QR, RS, SP and BD. BD is a diagonal of ABCD.

In

Similarly in

QR || BD and QR =

From equations (i) and (ii), we get:

SP || BD || QR

SP || QR and SP = OR [Each equal to

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

Therefore, SPQR is a parallelogram.

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

Therefore, PR and OS bisect each other.

**Q.13: In the given figure, ABCD is a quadrilateral whose diagonals intersect at right angles. Show that the quadrilateral formed by joining the midpoints of the pair of adjacent it is a rectangle.**

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**Sol:**

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

AC and BD are the diagonals which intersect each other at 0. RQ intersects AC at E and SR intersects BD at F.

In

PC || AC and PO = 12AC [By midpoint theorem]

Again, in ADAC, the points S and R are the midpoints of AD and DC, respectively. SR || AC and SR = 12AC [By midpoint theorem]

Now, PC || AC and SR || AC

PQ || SR

Also, PQ = SR [Each equal to 12AC]

So, PARS is a parallelogram.

We know that the diagonals of the given quadrilateral bisect each other at right angles.

∠EOF = 90°

Now, RQ || DB

i.e. RE || FO

Also, SR || AC

i.e. FR || OE

OERF is a parallelogram.

So, ∠FRE = ∠EOF = 90° (Opposite angles are equal)

Thus, PQRS is a parallelogram with ∠R = 90°.

PQRS is a rectangle.