##### Question 1:

The angles of a quadrilateral are in the ratio 3 : 5 : 9 : 13. Find all angles of the quadrilateral.

##### Answer:

The angles of a quadrilateral are in the ratio

3 : 5 : 9 : 13.

So, ∠A = 3x

∠B = 5x

∠C = 9x

∠D = 13x

where x is a positive constant.

Now, ∠A + ∠B + ∠C + ∠D = 360$\xb0$

[Angle sum property of a quadrilateral]

$\Rightarrow $ 3x + 5x + 9x + 13x = 360$\xb0$

$\Rightarrow $ 30x = 360$\xb0$

$\Rightarrow $ x = 12$\xb0$

Now,

∠A = 3x $\Rightarrow $ ∠A = 3 × 12$\xb0$ = 36$\xb0$

∠B = 5x $\Rightarrow $ ∠B = 5 × 12$\xb0$ = 60$\xb0$

∠C = 9x $\Rightarrow $ ∠C = 9 × 12$\xb0$ = 108$\xb0$

and ∠D = 13x $\Rightarrow $ ∠D = 13 × 12$\xb0$ = 156$\xb0$

Hence, angles of given quadrilateral are 36$\xb0$, 60$\xb0$, 108$\xb0$ and 156$\xb0$ respectively.

##### Question 2:

If the diagonals of a parallelogram are equal, show that it is a rectangle.

##### Answer:

Given: ABCD is a gm with diagonal AC

= diagonal BD

To prove: ABCD is a rectangle.

Proof: In $\u2206$s ABC and ABD, we have

AB = AB ... (Common side)

AC = BD (Given)

and AD = BC (Opp. sides of a II gm)

$\therefore $ $\u2206$ABC $\cong $ $\u2206$BAD

(By SSS congruence rule)

$\Rightarrow $ ∠DAB = ∠CBA ...(i)

(Property of congruency)

But ∠DAB + ∠CBA = 180$\xb0$ ...(ii)

[$\because $ AD II BC and AB cuts them, the sum of the

int. ∠s on the same side of transversal is 180$\xb0$.]

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

∠DAB = ∠CBA = 90$\xb0$

Hence, ABCD is a rectangle.

[$\because $ If one angle of II gm is 90$\xb0$, it is a rectangle.]

[Hence Proved]

##### Question 3:

Show that if diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus.

##### Answer:

Let ABCD be a rhombus.

Let its diagonal AC and BD bisect each other at right angle at point O.

$\therefore $ OA = OC, OB = OD

and ∠AOB = ∠BOC = ∠COD

= ∠AOD = 90$\xb0$

We have to prove that ABCD is a rhombus.

In $\u2206$AOD and $\u2206$BOC,

OA = OC (Given)

∠AOD = ∠BOC (Each = 90$\xb0$) (Given)

OD = OB (Given)

$\therefore $ $\u2206$AOD $\cong $ $\u2206$ COB

(SAS criteria of congruency)

So, AD = CB ...(i)

(Corresponding parts of congruent triangles)

In $\u2206$AOB and $\u2206$COD,

OA = OC (Given)

∠AOB = ∠COD (Each = 90$\xb0$) (Given)

OB = OD (Given)

$\therefore $ $\u2206$AOB $\cong $ $\u2206$COD

(SAS criteria of congruency)

So, AB = CD (Corresponding part of

congruent triangles) ...(ii)

Now in $\u2206$AOB and $\u2206$BOC,

AO = OC (Given)

∠AOB = ∠BOC

(each = 90$\xb0$) (Given)

OB = OB (Common)

$\therefore $ $\u2206$AOB $\cong $ $\u2206$BOC

(SAS criteria of congruency)

So, AB = BC ...(iii)

(Corresponding parts of congruent triangles)

From (i), (ii) and (iii), we get

AD = BC = CD = AB

So, in addition to given conditions that the diagonals of a quadrilateral bisect each other
at right angle; we also have all its sides equal. Hence, quadrilateral satisfies all the
conditions to be a rhombus. Therefore given quadrilateral is a rhombus.

##### Question 4:

Show that the diagonals of a square are equal and bisect each other at right angles.

##### Answer:

Given: ABCD is a square. AC and BD are its diagonals bisect each other at point O.

To prove: (i) AC = BD, (ii) AC $\perp $ BD at point O.

<img height="218" src= "images/MBD_SR_METH_G9_img_300.jpg" width="360"> Proof: In $\u2206$s ABC and BAD,

AB = AB (Common side)

∠ABC = ∠BAD (Each 90$\xb0$)

BC = AD (Sides of a square)

$\therefore $ $\u2206$ABC $\cong $ $\u2206$BAD [SAS congruence rule]

$\Rightarrow $ AC = BD (c.p.c.t.)

Hence, part (i) is proved.

In $\u2206$s AOB and AOD,

AO = AO (Common sides)

AB = AD (Sides of a square)

OB = OD

(Diagonals of a square bisect each other)

$\therefore $ $\u2206$AOB $\cong $ $\u2206$AOD (SSS congruence rule)

$\therefore $ ∠AOB = ∠AOD (c.p.c.t.)

But ∠AOB + ∠AOD = 180$\xb0$ (Linear pair angles)

$\therefore $∠AOB = ∠AOD = 90$\xb0$

i.e. OA $\perp $ BD or AC $\perp $ BD.

Hence, part (ii) is proved.

##### Question 5:

Show that if the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square.

##### Answer:

Let ABCD be a quadrilateral in which equal diagonals AC and BD bisect each other at right
angle at point O.

We have

AC = BD

OA = OC ...(i)

and OB = OD ...(ii)

AC = BD

$\Rightarrow $ OA + OC = OB + OD

$\Rightarrow $ OC + OC = OB + OB [Using (i) and (ii)]

$\Rightarrow $ 2OC = 2OB

$\Rightarrow $ OC = OB ...(iii)

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

OA = OB = OC = OD ...(iv)

Now in $\u2206$AOB and COD,

OA = OD [Shown in part (iv)]

∠AOB = ∠COD

[Vertically opposite angles]

OB = OC [Shown in part (iv)]

$\therefore $ $\u2206$AOB $\cong $ $\u2206$DOC

[SAS criteria of congruency]

So AB = DC (c.p.c.t) ...(v)

Similarly, $\u2206$BOC $\cong $ $\u2206$AOD

[SAS criteria of congruency]

So, BC = AD (c.p.c.t) ...(vi)

(v) and (vi) implies that opposite sides of

quadrilateral ABCD are equal.

Hence, ABCD is a parallelogram.

Now in $\u2206$ABC and $\u2206$BAD,

AB = BA [Common side]

BC = AD [Proved in part (vi)]

AC = BD (Given)

$\therefore $ $\u2206$ABC $\cong $ $\u2206$BAD

[SSS criteria of congruency]

So, ∠ABC = ∠BAD (c.p.c.t) ...(vii)

But ∠ABC + ∠BAD = 180$\xb0$ ...(viii)

[$\because $ ABCD is a parallelogram (Proved above)]

$\therefore $ AD || BC and AB as a transversal

$\Rightarrow $ ∠ABC + ∠ABC = 180$\xb0$ [Using (vii) in (viii)]

$\Rightarrow $ 2 ∠ABC = 180$\xb0$

$\Rightarrow $ ∠ABC = 90$\xb0$

$\therefore $ ∠ABC = ∠BAD = 90$\xb0$ ...(ix)

Opposite angles of a parallelogram are equal

But ∠ABC = 90$\xb0$ and ∠BAD = 90$\xb0$

$\therefore $ ∠ABC = ∠ADC = 90$\xb0$ ...(x)

and ∠BAD = ∠BCD = 90$\xb0$ ...(xi)

We observe that

∠ABC = ∠ADC = ∠BAD

= ∠BCD = 90$\xb0$ ...(xii)

Now in $\u2206$AOB and $\u2206$BOC

OA = OC (given)

∠AOB = ∠BOC [each 90$\xb0$ (given)]

OB = OB (Common)

$\therefore $ $\u2206$AOB $\cong $ $\u2206$COB

(SAS rule of congruency)

So, AB = BC ...(xiii)

From (v), (vi) and (xiii), we get:

AB = BC = CD = AD ...(xiv)

Using (xii) and (xiv);

Now we have the quadrilateral whose equal diagonals bisect each other at right angle.

Also sides are equal making an angle of 90$\xb0$ with each other.

Hence, given quadrilateral satisfy all conditions to be a square.

##### Question 6:

Diagonal AC of a parallelogram ABCD bisects ∠A (see fig.). Show that

(i) It bisects ∠C also

(ii) ABCD is a rhombus.

##### Answer:

It is given that diagonal AC bisects ∠A of the || gm ABCD.

To prove: AC bisects ∠C

Proof: Since AB DC and AC intersects them

$\therefore $ ∠1 = ∠3 (Alternate angles) ...(a)

Similarly, ∠2 = ∠4 ...(b)

But ∠1 = ∠2 ($\because $ AC bisects ∠A) ...(c)

∠3 = ∠4 [Using (a), (b) and (c)]

Thus, AC bisects ∠C.

##### Question 7:

ABCD is a rhombus. Show that the diagonal AC bisects ∠A as well as ∠C and diagonal BD bisects ∠B as well as ∠D.

##### Answer:

ABCD is a rhombus.

$\therefore $ AB = BC = CD = AD

Let O be the point of bisection of diagonals.

$\therefore $ OA = OC and OB = OD

In $\u2206$AOB and $\u2206$AOD,

OA = OA (Common)

AB = AD[Equal sides of rhombus]

OB = OD

[Diagonals of rhombus bisect each other]

$\therefore $ $\u2206$AOB $\cong $ $\u2206$AOD

[SSS criteria of congruency]

So, ∠OAD = ∠OAB

(Corresponding parts of congruent triangles)

$\Rightarrow $ OA bisects ∠A ....(i)

Similarly, $\u2206$BOC $\cong $ $\u2206$DOC

(SSS criteria of congruency)

So; ∠OCB = ∠OCD [c.p.c.t]

$\Rightarrow $ OC bisects ∠C ...(ii)

From (i) and (ii), we can say that diagonal AC

bisects ∠A and ∠C.

Now in $\u2206$AOB and $\u2206$BOC,

OB = OB (Common)

AB = BC [Equal sides of rhombus]

OA = OC

[$\because $ Diagonals of rhombus bisect each other]

$\therefore $ $\u2206$AOB $\cong $ $\u2206$COB

[SSS criteria of congruency]

So, ∠OBA = ∠OBC (c.p.c.t)

$\Rightarrow $ OB bisects ∠B ...(iii)

Similarly, $\u2206$AOD $\cong $ $\u2206$COD [SSS congruency]

$\Rightarrow $ ∠ODA = ∠ODC (c.p.c.t.)

From (iii) and (iv) we can say that diagonal BD bisects ∠B and ∠D.

##### Question 8:

ABCD is a rectangle in which diagonal AC bisects ∠A as well as ∠C. Show that:

ABCD is a square

##### Answer:

ABCD is a rectangle

$\therefore $ AB = DC ...(a)

and BC = AD

Also each angle; ∠A = ∠B = ∠C = ∠D = 90$\xb0$

In $\u2206$ABC and $\u2206$ADC,

∠1 = ∠2

and ∠3 = ∠4

[$\because $ AC bisects ∠A and ∠C (given)]

AC = AC (Common)

$\therefore $ $\u2206$ABC $\cong $ $\u2206$ADC

(ASA criteria of congruency)

So, AB = AD ...(b)

From (a) and (b), we get:

AB = BC = AD = DC

It implies that all sides of a rectangle are

equal.

Hence, it is a square.

##### Question 9:

ABCD is a rectangle in which diagonal AC bisects ∠A as well as ∠C. Show that:

Diagonal BD bisects both ∠B as well as ∠D.

##### Answer:

ABCD is a rectangle

$\therefore $ AB = DC ...(a)

and BC = AD

Also each angle; ∠A = ∠B = ∠C = ∠D = 90$\xb0$

In $\u2206$ABD and $\u2206$BDC

AB = BC [$\because $ Rectangle ABCD is a square proved in part (i)]

AD = DC

(proved in part (i) as ABCD is a square)

BD = BD (Common)

$\therefore $ $\u2206$ABD $\cong $ $\u2206$CBD

(SSS criteria of congruency)

So, ∠ABD = ∠CBD (c.p.c.t.) ...(c)

and ∠ADB = ∠CDB (c.p.c.t.) ...(d)

(c) and (d) implies that diagonal BD bisects

both ∠B and ∠D.

##### Question 10:

In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP = BQ (see fig.). Show that

$\u2206$APD $\cong $ $\u2206$CQB

##### Answer:

In $\u2206$APD and $\u2206$CQB,

DP = BQ (given)

∠ADP = ∠QBC

[$\because $ In parallelogram ABCD,

AD BC; BD being a transversal.

$\therefore $ ∠ADB = ∠DBC (alternate angles)

So;∠ADP = ∠QBC]

AD = CB [$\because $ Opposite sides of parallelogram are equal]

$\therefore $ $\u2206$APD $\cong $ $\u2206$CQB (SAS criteria of congruency)

##### Question 11:

In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP = BQ (see fig.). Show that

AP = CQ

##### Answer:

So; AP = CQ (Corresponding parts of congruent triangles)

##### Question 12:

In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP = BQ (see fig.). Show that

$\u2206$AQB $\cong $ $\u2206$CPD

##### Answer:

In $\u2206$AQB and $\u2206$CPD,

BQ = DP (Given)

∠ABQ = ∠PDC

[$\because $ In parallelogram ABCD,

AB || CD; BD being a transversal.

$\therefore $ ∠ABD = ∠BDC (Alternate angles)

So;∠ABQ = ∠PDC]

AB = CD [$\because $Opposite sides of a parallelogram are equal]

$\therefore $ $\u2206$AQB $\cong $ $\u2206$CPD

[SAS criteria of congruency]

##### Question 13:

AQ = CP

##### Answer:

So, AQ = CP (Corresponding parts of congruent triangles)

##### Question 14:

APCQ is a parallelogram

##### Answer:

In quadrilateral APCQ; we have

AP = CQ [Proved in part (ii)]

AQ = CP [Proved in part (iv)]

Opposite sides of quadrilateral APCQ are equal.

As we know that in a parallelogram opposite sides are equal.

Hence, APCQ is a parallelogram.

##### Question 15:

ABCD is a parallelogram and AP and CQ are the perpendiculars from vertices A and C on its
diagonal BD (see fig.).

Show that:

$\u2206$APB $\cong $ $\u2206$CQD

##### Answer:

ABCD is a parallelogram

$\therefore $ AB DC

BD being a transversal.

So, ∠1 = ∠2 (Alternate angles)

Now in $\u2206$APB and $\u2206$CQD,

∠APB = ∠CQD (each = 90$\xb0$) (Given)

∠1 = ∠2 (Proved above)

AB = CD [$\because $Opposite sides of a

parallelogram are equal]

$\therefore $ $\u2206$APB $\cong $ $\u2206$CQD

(AAS criteria of congruency)

##### Question 16:

ABCD is a parallelogram and AP and CQ are the perpendiculars from vertices A and C on its
diagonal BD (see fig.).

Show that:

AP = CQ

##### Answer:

ABCD is a parallelogram

$\therefore $ AB DC

BD being a transversal.

So, ∠1 = ∠2 (Alternate angles)

So, AP = CQ [Corresponding parts of congruent triangles]

##### Question 17:

In $\u2206$ABC and $\u2206$DEF; AB = DE, AB II DE, BC = EF and BC II EF. Vertices A, B and C are joined to
vertices D, E and F respectively (see fig.).

Show that

Quadrilateral ABED is a parallelogram

##### Answer:

Given: In $\u2206$ABC and $\u2206$DEF,

AB = DE and AB II DE

Also in $\u2206$s, BC = EF and BC II EF

In quad. ABED, its one pair of opposite sides AB and DE are such that AB = DE and AB II
DE

$\therefore $ ABED is a II gm.

$\therefore $ AD = BE and AD II BE.

[Opp. sides of a gm are equal and parallel] ... (i)

##### Question 18:

In $\u2206$ABC and $\u2206$DEF; AB = DE, AB II DE, BC = EF and BC II EF. Vertices A, B and C are joined to
vertices D, E and F respectively (see fig.).

Show that

Quadrilateral BEFC is a parallelogram

##### Answer:

Given: In $\u2206$ABC and $\u2206$DEF,

AB = DE and AB II DE

Also in $\u2206$s, BC = EF and BC II EF

Again, quad. BEFC

BE = CF and BE CF

$\therefore $ BEFC is a gm.

##### Question 19:

In $\u2206$ABC and $\u2206$DEF; AB = DE, AB DE, BC = EF and BC EF. Vertices A, B and C are joined to vertices D,
E and F respectively (see fig.).

Show that

AD II CF and AD = CF

##### Answer:

Given: In $\u2206$ABC and $\u2206$DEF,

AB = DE and AB II DE

Also in $\u2206$s, BC = EF and BC EF

$\therefore $ CF = BE and CF II BE ... (ii)

From (1) and (2), we get

AD = CF and AD II CF

##### Question 20:

In $\u2206$ABC and $\u2206$DEF; AB = DE, AB II DE, BC = EF and BC II EF. Vertices A, B and C are joined to
vertices D, E and F respectively (see fig.).

Show that

Quadrilateral ACFD is a parallelogram

##### Answer:

Given: In $\u2206$ABC and $\u2206$DEF,

AB = DE and AB II DE

Also in $\u2206$s, BC = EF and BC II EF

$\therefore $ ACFD is a IIgm. [$\because $If one pair of opp. sides

of quad. is equal and parallel, it is a IIgm.]

##### Question 21:

Show that

AC = DF

##### Answer:

Given: In $\u2206$ABC and $\u2206$DEF,

AB = DE and AB II DE

Also in $\u2206$s, BC = EF and BC II EF

Hence, AC = DF [Opp. sides of a gm]

##### Question 22:

In $\u2206$ABC and $\u2206$DEF; AB = DE, AB II DE, BC = EF and BC II EF. Vertices A, B and C are joined to vertices D, E and F respectively (see fig.).

Show that$\u2206$ABC $\cong $ $\u2206$DEF

##### Answer:

Given: In $\u2206$ABC and $\u2206$DEF,

AB = DE and AB II DE

Also in $\u2206$s, BC = EF and BC II EF

In $\u2206$ABC and $\u2206$DEF,

AB = DE (Given)

BC = EF (Given)

AC = DF [Proved in part (v)]

$\therefore $ $\u2206$ABC $\cong $ $\u2206$DEF

(SSS criteria of congruency)

##### Question 23:

ABCD is a trapezium in which AB II CD and AD = BC (See fig.)

Show that:

∠A = ∠B

##### Answer:

Extend AB and draw a line CE parallel to AD as shown in the figure.

Since, AD II CE and transversal AE cuts them at A and E respectively.

$\therefore $ ∠A + ∠E = 180$\xb0$

$\Rightarrow $ ∠A = 180$\xb0$ – ∠E ...(i)

Since AB II CD and AD II CE

$\therefore $ AECD is a IIgm

$\Rightarrow $ AD = CE

$\Rightarrow $ BC = CE [$\because $AD = BC (Given)]

Thus, in $\u2206$BCE, we have

BC = CE

$\Rightarrow $ ∠CBE = ∠CEB

[Equal angles opp. to equal sides]

$\Rightarrow $ 180$\xb0$ – ∠B = ∠E

[$\because $ ∠CBE + ∠ABC = 180$\xb0$ (Linear pair)

$\therefore $ ∠CBE = 180$\xb0$ – ∠ABC]

$\Rightarrow $ 180$\xb0$ – ∠E = ∠B ...(ii)

From (i) and (ii), we get ∠A = ∠B

##### Question 24:

ABCD is a trapezium in which AB CD and AD = BC (See fig.)

Show that:

∠C = ∠D

##### Answer:

ABCD is a trapezium in which AB || DC

$\therefore $ ∠A + ∠D = 180$\xb0$ ...(a)

and ∠B + ∠C = 180$\xb0$ ...(b)

[$\because $For two parallel lines sum of interior angles on the same side of a transversal is
180$\xb0$.]

Equating (a) and (b), we get:

∠A + ∠D = ∠B + ∠C

But ∠A = ∠B proved in part (i)

$\therefore $ ∠A + ∠D = ∠A + ∠C

$\Rightarrow $ ∠D = ∠C

or ∠C = ∠D [Hence proved]

##### Question 25:

ABCD is a trapezium in which AB CD and AD = BC (See fig.)

Show that:

$\u2206$ABC $\cong $ $\u2206$BAD

##### Answer:

In $\u2206$ABC and $\u2206$BAD,

AB = AB, (Common)

∠A = ∠B [Proved in part (i)]

BC = AD (given)

$\therefore $ $\u2206$ABC $\cong $ $\u2206$BAD

(SAS criteria of congruency)

##### Question 26:

ABCD is a trapezium in which AB CD and AD = BC (See fig.)

Show that:

diagonal AC = diagonal BD.

##### Answer:

So, AC = BD (Corresponding part of congruent triangles)

i.e., in trapezium ABCD;

diagonal AC = diagonal BD.

##### Question 27:

ABCD is a quadrilateral in which P, Q, R and S are the mid-points of sides AB, BC, CD and DA respectively (see fig.)

AC is a diagonal.

Show that

##### Answer:

In $\u2206$ABC,

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

Then PQ II AC

In $\u2206$ACD,

R is the mid-point of CD and S is the

mid-point of AD.

##### Question 28:

ABCD is a quadrilateral in which P, Q, R and S are the mid-points of sides AB, BC, CD and DA respectively (see fig.)

AC is a diagonal.

Show that

PQ = SR

##### Answer:

In $\u2206$ABC,

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

Then PQ AC

##### Question 29:

ABCD is a quadrilateral in which P, Q, R and S are the mid-points of sides AB, BC, CD and DA respectively (see fig.)

AC is a diagonal.

Show that

PQRS is a parallelogram.

##### Answer:

In $\u2206$ABC,

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

Then PQ II AC

PQ II AC

and SR II AC

$\Rightarrow $ PQ II SR [$\because $ Two lines parallel

to a given line are parallel to each other.]

Now, we have

PQ = SR

and PQ II SR

As we know that if one pair of opposite sides of a quadrilateral is equal and parallel then it is a parallelogram.

$\therefore $ PQRS is a parallelogram.

##### Question 30:

ABCD is a rhombus and P, Q, R, S are the mid-points of AB, BC, CD and DA respectively. Prove that quadrilateral PQRS is a rectangle.

##### Answer:

Given: P, Q, R and S are the mid-points of respective sides AB, BC, CD and DA of rhombus.

PQ, QR, RS and SP are joined.

To prove: PQRS is a rectangle.

Construction: Join A and C.

Proof: In $\u2206$ABC, P is the mid-point of AB and Q is the mid-point of BC.

$\therefore $ By mid-point theorem,

Now in $\u2206$s APS and CQR, we have AP = CQ

$\Rightarrow $ AP = CQ where P and Q are midpoints

of AB and BC.]

Similarly, AS = CR

PS = QR

[Opposite sides of gm PQRS]

$\therefore $ $\u2206$APS $\cong $ $\u2206$CQR

[By SSS Congruence rule]

$\therefore $ ∠3 = ∠4 (c.p.c.t.)

Now we have

∠1 + ∠SPQ + ∠3 = 180$\xb0$

and ∠2 + ∠PQR + ∠4 = 180$\xb0$ [Linear pairs]

$\therefore $ ∠1 + ∠SPQ + ∠3 = ∠2 + ∠PQR + ∠4

Since ∠1 = ∠2 and ∠3 = ∠4

(As proved above)

$\therefore $ ∠SPQ = ∠PQR ... (iii)

Now, PQRS is a gm (as proved above)

$\therefore $ ∠SPQ + ∠PQR = 180$\xb0$ ... (iv)

[$\because $ SP RQ and PQ cuts them and the sum

of int. ∠s on the same side of a transversal is

180$\xb0$.]

Using (iii) in (iv) we get:

∠SPQ + ∠SPQ = 180$\xb0$

$\Rightarrow $ 2∠SPQ = 180$\xb0$

$\Rightarrow $ ∠SPQ = 90$\xb0$

Thus, PQRS is a gm whose one angle

∠SPQ = 90$\xb0$.

Hence, PQRS is a rectangle.

##### Question 31:

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.

##### Answer:

Given: A rectangle ABCD in which P, Q, R and S are the mid-points of sides AB, BC, CD and DA respectively PQ, QR, RS and SP are joined.

To prove: PQRS is a rhombus

Construction: Join AC

Proof: In $\u2206$ABC, P and Q are the mid-points of sides AB, BC respectively.

$\Rightarrow $ AS = BQ ... (iv)

In $\u2206$s APS and BPQ, we have:

AP = BP

[$\because $ P is the mid-point of AB.]

∠PAS = ∠PBQ [Each is 90$\xb0$]

and AS = BQ [From (iv)]

$\therefore $ $\u2206$APS $\cong $ $\u2206$BPQ

[SAS criterion of congruency]

$\Rightarrow $ PS = PQ ... (v)

[Corresponding parts of congruent triangles are equal]

From (iii) and (v), we get PQRS is a

parallelogram such that

PS = PQ

i.e., two adjacent sides are equal.

Hence, PQRS is a rhombus.

##### Question 32:

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

##### Answer:

Since E and F are the mid-points of AB and CD respectively.

$\Rightarrow $ AB = CD and AB DC$\Rightarrow $ AE = FC and AE II FC [From (i)]

$\Rightarrow $ AECF is a IIgm

$\Rightarrow $ FA II CE

$\Rightarrow $ FP II CQ ...(ii)

[$\because $ FP is a part of FA and CQ is a part of CE]

We know that, the segment drawn through the mid-point of one side of a triangle and parallel
to the other side bisects the third side.

In $\u2206$DCQ, F is the mid-point of CD and

FP II CQ [From (ii)]

$\therefore $ P is the mid-point of DQ

$\Rightarrow $ DP = PQ ...(iii)

Similarly, in $\u2206$ABP, E is the mid-point of AB

and EQ II AP

$\therefore $ Q is the mid-point of BP

$\Rightarrow $ BQ = PQ ...(iv)

From (iii) and (iv), we have

DP = PQ = BQ ...(v)

Now BD = BQ + PQ + DP

= BQ + BQ + BQ

$\Rightarrow $ BD = 3BQ

or 3BQ = BD

It implies that points P and Q trisects BD.

So, AF and CE trisects BD.

##### Question 33:

Show that the line segments joining the midpoints of opposite sides of a quadrilateral bisect each other.

##### Answer:

A quadrilateral ABCD in which EG and FH are the line-segments joining the mid-points of opposite sides of a quadrilateral.

To Prove: EG and FH bisect each other.

Construction: Join AC, EF, FG, GH and HE

Proof: In $\u2206$ABC, E and F are the mid-points of respective sides AB and BC.

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

EF HG and EF = HG

$\therefore $ EFGH is a parallelogram.

[$\because $ A quadrilateral is a parallelogram if one pair of its opposite sides is equal and
parallel.]

As we know, that the diagonals of a parallelogram bisect each other, therefore line segments
(i.e. diagonals) EG and FH (of parallelogram EFGH) bisect each other.

##### Question 34:

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:

D is the mid-point of AC

##### Answer:

In $\u2206$ABC, M is the mid-point of AB. (Given) MD II BC $\therefore $ AD = DC[Converse of Mid-point theorem] Thus, D is the mid-point of AC.##### Question 35:

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:

MD $\perp $ AC

##### Answer:

l II BC (Given)

Consider AC as a transversal

$\therefore $ ∠1 = ∠C(Corresponding angles)

$\Rightarrow $ ∠1 = 90$\xb0$ [$\because $ ∠C = 90$\xb0$ (Given)]

Thus, MD $\perp $ AC.

##### Question 36:

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:

##### Answer:

In $\u2206$AMD and $\u2206$CMD,

AD = DC (Proved above)

∠1 = ∠2 (Each = 90$\xb0$)

[Proved above]

MD = MD (Common)

$\u2206$AMD $\cong $ $\u2206$CMD

(SAS criteria of congruency)

So, AM = CM ...(a) (c.p.c.t.)

Given that M is the mid-point of AB.