# NCERT solution for class 9 maths quadrilaterals ( Chapter 8) #### Solution for Exercise 8.1

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

Given: The ratios of the angles of quadrilateral are 3: 5: 9: 13.
Let the angles of the quadrilateral be 3x, 5x, 9x and 13x.
We know that, sum of angles of a quadrilateral = $$360^\circ$$
i.e., 3x + 5x + 9x + 13x = $$360^\circ$$
i.e., 30x = $$360^\circ$$
Therefore, x = $$12^\circ$$

Thus, we get,
=> 3x = 3 × $$12^\circ$$ = $$36^\circ$$
=>5x = 5 × $$12^\circ$$ = $$60^\circ$$
=> 9x = 9 × $$12^\circ$$ = $$108^\circ$$
=> 13x = 13 × $$12^\circ$$ = $$156^\circ$$

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

Given: Parallelogram is ABCD whose diagonals AC and BD are equal. i.e., AC = BD

To prove: ABCD is a rectangle.

Proof: In $$\triangle{ABC}$$ and $$\triangle{DCB}$$, we have,
AC = BD ...(Given)
AB = CD ...(Opposite sides of parallelogram)
BC = CB ...(Common sides)
Therefore, $$\triangle{ABC}$$$$\displaystyle \cong$$$$\triangle{DCB}$$ ...(By SSS rule)
Thus, $$\angle{ABC}$$ = $$\angle{DCB}$$ ...(i)(By CPCT)

But from figure, DC || AB and transversal CB intersects them
$$\angle{ABC}$$ + $$\angle{DCB}$$ = $$180^\circ$$ ...(interior angles on same side of transversal)
$$\angle{ABC}$$ + $$\angle{ABC}$$ = $$180^\circ$$ ...(from(i))
i.e., 2 $$\angle{ABC}$$ = $$180^\circ$$
i.e., $$\angle{ABC}$$ = $$90^\circ$$
Also, $$\angle{DCB}$$ = $$90^\circ$$
Thus, we can say that, ABCD is a parallelogram and some angles are $$90^\circ$$.
Hence it is proved that, ABCD is a rectangle.

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

Given: A quadrilateral ABCD whose diagonals AC and BD bisect each other at right angles.
i.e., OA = OC and OB = OD
Also, $$\angle{AOD}$$ = $$\angle{AOB}$$ = $$\angle{COD}$$ = $$\angle{BOC}$$ = $$90^\circ$$. To prove: ABCD is a rhombus.

Proof: In $$\triangle{OAB}$$ and $$\triangle{ODC}$$, we have,
OA = OC and OB = OD ...(given)
$$\angle{AOB}$$ = $$\angle{COD}$$ ...(vertically opposite angles)
Therefore, $$\triangle{OAB}$$ $$\displaystyle \cong$$ $$\triangle{OCD}$$ ...(By SAS rule)
Thus, AB = CD ...(i)(By CPCT)

Similarly, in $$\triangle{OAD}$$ and $$\triangle{OBC}$$, we have,
OA = OC and OD = OB ...(given)
$$\angle{AOD}$$ = $$\angle{BOC}$$ ...(vertically opposite angles)
Therefore, $$\triangle{OAD}$$ $$\displaystyle \cong$$ $$\triangle{OCB}$$ ...(By SAS rule)
Thus, AD = BC ...(ii)(By CPCT)
Similarly, we can prove that,
AB = CD and
CD = BC ...(iii)
Hence, from (i), (ii) and (iii), we get,
AB = BC = AD = CD
Hence, it is proved that ABCD is a rhombus.

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

Given: A square ABCD whose diagonals AC and BD intersect at O.

To prove: Diagonals are equal and bisect each other at right angles. i.e, AC = BD, OD = OB, OA = OC and $${AC}\perp{BD}$$

Proof: In $$\triangle{ABC}$$ and $$\triangle{BAD}$$, we have,
AB = BA ...(Common side)
$$\angle{ABC}$$ = $$\angle{BAD}$$ = $$90^\circ$$
Therefore, $$\triangle{ABC}$$ $$\displaystyle \cong$$ $$\triangle{BAD}$$ ...(By SAS rule)
Thus, AC = BD ...(By CPCT)

Similarly, in $$\triangle{OAB}$$ and $$\triangle{OCD}$$, we have,
AB = DC ...(given)
$$\angle{OAB}$$ = $$\angle{DCO}$$ ...(Since, AB || CD and transversal AC intersect)
$$\angle{OBA}$$ = $$\angle{BDC}$$ ...(Since, AB || CD and transversal BD intersect)
Therefore, $$\triangle{OAB}$$ $$\displaystyle \cong$$ $$\triangle{OCD}$$ ...(By SAS rule)
Thus, OA = OC and OB = OD ...(By CPCT)

Now, in $$\triangle{AOB}$$ and $$\triangle{AOD}$$, we have,
OA = OD ...(proved earlier)
AB = AD ...(sides of square)
AO = OA ...(Commom side)
Therefore, $$\triangle{AOB}$$ $$\displaystyle \cong$$ $$\triangle{AOD}$$ ...(By SSS rule)
Thus, $$\angle{AOB}$$ = $$\angle{AOD}$$ ...(By CPCT)

$$\angle{AOB}$$ + $$\angle{AOD}$$ = $$180^\circ$$ ...(linear pair axiom)
$$\angle{AOB}$$ = $$\angle{AOD}$$ = $$90^\circ$$
Thus, $$AO\perp{BD}$$, i.e., $$AC\perp{BD}$$.
Also, AC = BD, OA = OC, OB = OD and $$AC\perp{BD}$$
Hence, it is proved that diagonals are equal and bisect each other at right angles.

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

Given: A quadrilateral ABCD in which AC = BD and AC $$\perp{BD}$$ such that OA = OC and OB = OD. So, ABCD is a parallelogram. To prove: ABCD is a square.

Proof: Let AC and BD intersect at a point O.
In $$\triangle{ABO}$$ and $$\triangle{ADO}$$, we have,
BO = OD ...(given)
AO = OA ...(Common side)
$$\angle{AOB}$$ = $$\angle{AOD}$$ = $$90^\circ$$ ...(given)
Therefore, $$\triangle{ABO}$$ $$\displaystyle \cong$$ $$\triangle{ADO}$$ ...(By SAS rule)
Thus, AB = AD ...(By CPCT)
Also, AB = DC and AD = BC ...(opposite sides of parallelogram)
Therefore, AB = BC = DC = AD ...(i)

Similarly, in $$\triangle{ABC}$$ and $$\triangle{BAD}$$, we have,
AC = BD ...(given)
AB = BA ...(Common side)
Therefore, $$\triangle{ABC}$$ $$\displaystyle \cong$$ $$\triangle{BAD}$$ ...(By SSS rule)
Thus, $$\angle{ABC}$$ = $$\angle{BAD}$$ ...(ii)(By CPCT)

But $$\angle{ABC}$$ + $$\angle{BAD}$$ = $$180^\circ$$ ...(linear pair axiom)
$$\angle{ABC}$$ = $$\angle{BAD}$$ = $$90^\circ$$ ...(from (ii))
Thus, AB = BC = CD = DA, and $$\angle{A}$$ = $$90^\circ$$
Hence, it is proved that, ABCD is a square.

6. Diagonal AC of a parallelogram ABCD bisects $$\angle{A}$$ (see figure). Show that
(i) It bisects $$\angle{C}$$ also,
(ii) ABCD is a rhombus. Given, diagonal AC of a parallelogram ABCD bisects $$\angle{A}$$. i.e., $$\angle{DAC}$$ = $$\angle{BAC}$$ = 1/2 $$\angle{BAD}$$ ...(i)
Here, AB || CD and AC is the transversal.
$$\angle{DCA}$$ = $$\angle{CAB}$$ ...(ii)(alternate angles)
$$\angle{BCA}$$ = $$\angle{DAC}$$ ...(iii)(alternate angles)

From eq. (i), (ii) and (iii), we get,
$$\angle{DAC}$$ = $$\angle{BAC}$$ = $$\angle{DCA}$$ = $$\angle{BCA}$$
Now, $$\angle{BCD}$$ = $$\angle{BCA}$$ + $$\angle{DCA}$$
i.e., = $$\angle{DAC}$$ + $$\angle{CAB}$$
Therefore, $$\angle{BCD}$$ = $$\angle{BAD}$$
Thus, we can say that, diagonal AC bisects $$\angle{C}$$.

Now, in $$\triangle{OAD}$$ and $$\triangle{OCD}$$, we have,
OA = OC ...(since, diagonals bisects each other)
DO = OD ...(Common side)
$$\angle{AOD}$$ = $$\angle{COD}$$ = $$90^\circ$$
Therefore, $$\triangle{OAD}$$ $$\displaystyle \cong$$ $$\triangle{OCD}$$ ...(By SAS rule)
Thus, AD = CD ...(By CPCT)

Now, AB = CD and AD = BC ...(sides of parallelogram)
Therefore, AB = CD = AD = BC
Hence it is proved that, ABCD is a rhombus.

7. ABCD is a rhombus. Show that diagonal AC bisects $$\angle{A}$$ as well as $$\angle{C}$$ and diagonal BD bisects $$\angle{B}$$ as well as $$\angle{D}$$.

Given: ABCD is a rhombus. i.e., AD = AB = BC = CD ... (i) To prove: (i) Diagonal AC bisect $$\angle{A}$$ as well as $$\angle{C}$$.
(ii) Diagonal BD bisects $$\angle{B}$$ as well as $$\angle{D}$$.

Proof: Let AC and BD are the diagonals of rhombus ABCD.
In $$\triangle{ABC}$$ and $$\triangle{ADC}$$, we have,
AC = CA ...(Common side)
CD = BC ...(From eq. (i))
Therefore, $$\triangle{ABC}$$ $$\displaystyle \cong$$ $$\triangle{ADC}$$ ...(By SSS rule)
Thus, $$\angle{DAC}$$ = $$\angle{BAC}$$ ...(By CPCT)
Also, $$\angle{DCA}$$ = $$\angle{BCA}$$
Also, $$\angle{DAC}$$ = $$\angle{DCA}$$
And $$\angle{BAC}$$ = $$\angle{BCA}$$
This shows that, Diagonal AC bisect $$\angle{A}$$ as well as $$\angle{C}$$.

Now, in $$\triangle{BDC}$$ and $$\triangle{BDA}$$, we have,
AB = BC ...(Given)
BD = BD ...(Common side)
Therefore, $$\triangle{BDC}$$ $$\displaystyle \cong$$ $$\triangle{BDA}$$ ...(By SSS rule)
Thus, $$\angle{BDA}$$ = $$\angle{BDC}$$ ...(By CPCT)
Also, $$\angle{DBA}$$ = $$\angle{DBC}$$
Also, $$\angle{BDA}$$ = $$\angle{DBA}$$
And $$\angle{BDC}$$ = $$\angle{DBC}$$
This shows that, Diagonal BD bisects $$\angle{B}$$ as well as $$\angle{D}$$.
Hence, proved.

8. ABCD is a rectangle in which diagonal AC bisects $$\angle{A}$$ as well as $$\angle{C}$$. Show that:
(i) ABCD is a square
(ii) Diagonal BD bisects $$\angle{B}$$ as well as $$\angle{D}$$.

Given: ABCD is a rectangle. AB = CD and BC = AD ...(i) To prove: (i) ABCD is a square. i.e., AB = BC = CD = DA
(ii) Diagonal BD bisects $$\angle{B}$$ as well as $$\angle{D}$$.

Proof: In $$\triangle{ADC}$$ and $$\triangle{ABC}$$, we have,
$$\angle{DAC}$$ = $$\angle{BAC}$$ ...(Since, AB || DC and AC is transversal that intersects)
Similarly, $$\angle{DCA}$$ = $$\angle{BCA}$$
AC = CA ...(Common side)
Therefore, $$\triangle{ADC}$$ $$\displaystyle \cong$$ $$\triangle{ABC}$$ ...(By ASA rule)
Thus, AD = AB ...(By CPCT)
Also, CD = BC ...(ii)
Thus, from eq. (i) and(ii), we get,
AB = BC = AD = CD
Hence, it is proved that ABCD is a square.

Now, in $$\triangle{AOB}$$ and $$\triangle{COB}$$, we have,
AB = BC ...(Given)
BO = OB ...(Common side)
OA = OC ...(Since,diagonal bisectS each other)
Therefore, $$\triangle{AOB}$$ $$\displaystyle \cong$$ $$\triangle{COB}$$ ...(By SSS rule)
Thus, $$\angle{OBA}$$ = $$\angle{OBC}$$ ...(By CPCT)
This shows that, Diagonal BD bisects $$\angle{B}$$.

Similarly, Now, in $$\triangle{AOD}$$ and $$\triangle{COD}$$, we have,
OD = DO ...(Common side)
OA = OC ...(Since,diagonal bisectS each other)
Therefore, $$\triangle{AOD}$$ $$\displaystyle \cong$$ $$\triangle{COD}$$ ...(By SSS rule)
Thus, $$\angle{ADO}$$ = $$\angle{CDO}$$ ...(By CPCT)
Hence, it is proved that, Diagonal BD bisects $$\angle{D}$$, too.

9. In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP = BQ (see figure). Show that
i) $$\triangle{APD}$$ $$\displaystyle \cong$$ $$\triangle{CQB}$$
ii) AP = CQ
iii) $$\triangle{AQB}$$ $$\displaystyle \cong$$ $$\triangle{CPD}$$
iv) AQ = CP
v) APCQ is a parallelogram. Given, ABCD is a parallelogram and P and Q are lie on BD such that DP = BQ ...(i)

i) Now,in $$\triangle{APD}$$ and $$\triangle{CQB}$$, we have,
DP = BQ ...(Given)
AD = BC ...(Opposite sides are equal in parallelogram)
$$\angle{ADP}$$ = $$\angle{QBC}$$ ...(Since, AD || BC and BD is a transversal)
Therefore, $$\triangle{APD}$$ $$\displaystyle \cong$$ $$\triangle{CQB}$$ ...(By SAS rule)

ii)Since, $$\triangle{APD}$$ $$\displaystyle \cong$$ $$\triangle{CQB}$$
Therefore, AP = CQ ...(By CPCT)

iii) Here, Now, in $$\triangle{AQB}$$ and $$\triangle{CPD}$$, we have,
DP = BQ ...(Given)
AB = CD ...(Opposite sides are equal in parallelogram)
$$\angle{ABQ}$$ = $$\angle{CDP}$$ ...(Since, AB || CD and BD is a transversal)
Therefore, $$\triangle{AQB}$$ $$\displaystyle \cong$$ $$\triangle{CPD}$$ ...(By SAS rule)

iv)Since, $$\triangle{AQB}$$ $$\displaystyle \cong$$ $$\triangle{CPD}$$
Therefore, AQ = CP...(By CPCT)

v) Now,in $$\triangle{APQ}$$ and $$\triangle{PCQ}$$, we have,
AQ = CP (From part iv))
AP = CQ (From part ii))
PQ = QP ...(Common side)
Therefore, $$\triangle{APQ}$$ $$\displaystyle \cong$$ $$\triangle{PCQ}$$ ...(By SSS rule)
Thus, $$\angle{APQ}$$ = $$\angle{PQC}$$
And $$\angle{AQP}$$ = $$\angle{CPQ}$$ ...(Vertically Opposite angles)

Now, these equal angles form a pair of alternate angle when line segment AP and QC are intersected by a transversal PQ.
i.e., AP || CQ and AQ || CP
Now, both pairs of opposite sides of quadrilateral APCQ are parallel.
Hence, it is proved that APCQ is a parallelogram.

10. ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD (see figure). Show that
(i) $$\triangle{APB}$$ $$\displaystyle \cong$$ $$\triangle{CQD}$$
(ii) AP = CQ Given: ABCD is a parallelogram and AP and CQ are perpendicular from vertices A and C on diagonal BD.

Since, AB || CD and BD is a transversal, we get,
$$\angle{CDB}$$ = $$\angle{DBA}$$ ...(i)
Now, in $$\triangle{APB}$$ and $$\triangle{CQD}$$, we have,
CD = AB ...(Sides of parallelogram)
$$\angle{CQD}$$ = $$\angle{APB}$$ = $$90^\circ$$ ...(Given)
$$\angle{CDQ}$$ = $$\angle{ABP}$$ ...(From Eq. (i))
Therefore, $$\triangle{APB}$$ $$\displaystyle \cong$$ $$\triangle{CQD}$$ ...(By ASA rule)
Thus, AP = CQ ...(By CPCT)
Hence, it is proved.

11. In $$\triangle{ABC}$$ and $$\triangle{DEF}$$, AB = DE, AB || DE, BC = EF and BC || EF. Vertices A, B and C are joined to vertices D, E and F, respectively (see figure). Show that
(i) Quadrilateral ABED is a parallelogram
(ii) Quadrilateral BEFC is a parallelogram
(iv) Quadrilateral ACFD is a Parallelogram
(v) AC = DF
(vi) $$\triangle{ABC}$$ $$\displaystyle \cong$$ $$\triangle{DEF}$$

Given: In $$\triangle{ABC}$$ and $$\triangle{DEF}$$, AB = DE, AB || DE, BC = EF and BC || EF

AB = DE and AB || DE ...(Given)
Since, a pair of opposite sides is equal and parallel
Therefore, ABED is a parallelogram.

BC = EF and BC || EF ...(Given)
Since, a pair of opposite sides is equal and parallel
Therefore, BEFC is a parallelogram.

(iii)Since, ABED is a parallelogram,
Also, BEFC in a parallelogram,
CF || BE and CF = BE ...(ii)
Thus, from Eq. (i) and (ii), we get,
Therefore, ACFD is a parallelogram.

(v)Since, ACFD is a parallelogram.
we get, AC = DF and AC || DF

(vi)Now, in $$\triangle{ABC}$$ and $$\triangle{DEF}$$,
AB = DE ...(Given)
BC = EF ...(Given)
and AC = DF ...(From part (v))
Therefore, $$\triangle{ABC}$$ $$\displaystyle \cong$$ $$\triangle{DEF}$$ ...(By SSS test)

12. ABCD is a trapezium in which AB || CD and AD = BC (see figure). Show that
i) $$\angle{A}$$ = $$\angle{B}$$
ii) $$\angle{C}$$ = $$\angle{D}$$
iii) $$\triangle{ABC}$$ $$\displaystyle \cong$$ $$\triangle{BAD}$$
iv) Diagonal AC = Diagonal BD
[Hint: Extend AB and draw a line through C parallel to DA intersecting AB produced at E].

Given: ABCD is trapezium.
AB || CD and AD = BC
Now, extend AB and draw a line through C parallel to DA intersecting AB produced at E. Therefore, AD = BC = CE

i) We know that, $$\angle{A}$$ + $$\angle{E}$$ = $$180^\circ$$ ...(Since, interior angles on the same side of the transversal )
i.e., $$\angle{E}$$ = $$180^\circ$$ - $$\angle{A}$$
Therefore, $$\angle{E}$$ = $$180^\circ$$ - $$\angle{A}$$ ...(Since, BC = EC)

Also, $$\angle{ABC}$$ = $$180^\circ$$ - $$\angle{CBE}$$ ...(Since, ABE is straight line)
$$\angle{ABC}$$ = $$180^\circ$$ - $$180^\circ$$ + $$\angle{A}$$
$$\angle{B}$$ = $$\angle{A}$$
Hence, it is proved.

ii)Now, $$\angle{A}$$ + $$\angle{D}$$ = $$180^\circ$$ ...(Since, interior angles on the same side of the transversal)
i.e., $$\angle{D}$$ = $$180^\circ$$ - $$\angle{A}$$
i.e., $$\angle{D}$$ = $$180^\circ$$ - $$\angle{B}$$ ...(from eq.(i)) ...(ii)

Also, $$\angle{C}$$ + $$\angle{B}$$ = $$180^\circ$$ ...180 (Since, interior angles on the same side of the transversal BC)
i.e., $$\angle{C}$$ = $$180^\circ$$ - $$\angle{B}$$ ...(iii)
from Eq. (ii) and (iii), we get,
$$\angle{C}$$ = $$\angle{D}$$
Hence, proved.

iii)Now, in $$\triangle{ABC}$$ and $$\triangle{BAD}$$, we have,
$$\angle{A}$$ = $$\angle{B}$$ ...(From Eq.(i))
AB = BA ...(Common side)
Therefore, $$\triangle{ABC}$$ $$\displaystyle \cong$$ $$\triangle{BAD}$$ ...(By SAS rule)
Hence, proved.

iv)Since, $$\triangle{ABC}$$ $$\displaystyle \cong$$ $$\triangle{BAD}$$
Therefore, AC = BD
Hence, it is proved.

#### Solution for Exercise 8.2

1. ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA (see figure). AC is a diagonal. Show that
i) SR || AC and SR = (1/2) AC
ii) PQ = SR
iii) PQRS is a parallelogram. Given: P, Q, R, and S are mid-points of the sides.
Therefore, AP = PB, BQ = CQ CR = DR and AS = DS

i) Now, in $$\triangle{ADC}$$, we have,
S is the midpoint of AD and R is the mid point of CD.
As, we know that,
By midpoint theorem, the line segment joining the mid points of two sides of a triangle is parallel to the third side.
Thus, we can say that, SR || AC ...(i)
and SR = (1/2) AC ...(ii)

ii) Similarly, now, in $$\triangle{ABC}$$, we have,
PQ || AC ...(iii)
and PQ = (1/2) AC ...(iv)
Now, from (ii) and (iv), we get,
SR = PQ = (1/2) AC ...(v)

Now, from (i), (iii) and (v), we get,
PQ || SR and PQ = SR.
Therefore, PQRS is a parallelogram ...(Since, a pair of opposite sides of a quadrilateral PQRS is equal and parallel)
Hence, proved.

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. Given: ABCD is a rhombus and P, Q, R and S are mid-points of AB, BC, CD and DA.

By midpoint theorm, Now, in $$\triangle{ADC}$$, we have,
Thus, SR || AC ...(i)
and SR = (1/2) AC ...(ii)
Similarly, in $$\triangle{ABC}$$, we have,
PQ || AC ...(iii)
and PQ = (1/2) AC ...(iv)
from (i), (ii), (iii) and (iv), we get,
SR = PQ = (1/2) AC and PQ || SR
Therefore, Quadrilateral PQRS is a parallelogram.

Now, we know that diagonals of a rhombus bisect each other at right angles.
Therefore, $$\angle{EOF}$$ = $$90^\circ$$.
Now, By midpoint theorem, we have,
RQ || BD
Thus, RE || OF
As SR || AC ...(from (i))
Thus, FR || OE
Therefore, OERF is a parallelogram
So, $$\angle{ERF}$$ = $$\angle{EOF}$$ = $$90^\circ$$ ...(since, Opposite angle of a quadrilateral is equal)
Thus, PQRS is a parallelogram with $$\angle{R}$$ = $$90^\circ$$
Hence, it is proved that PQRS is a rectangle.

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. Given: ABCD is a rectangle.
i.e., $$\angle{A}$$ = $$\angle{B}$$ = $$\angle{C}$$ = $$\angle{D}$$ = $$90^\circ$$ and AB = CD and BC = AD.
Also, P, Q, R and S are mid-points of the sides AB, BC, CD and DA, respectively.
Therefore, by midpoint theorem,
PQ || BD and PQ = (1/2) BD
and SR || AC and SR = (1/2) AC
In rectangle ABCD,
AC = BD
Thus, PQ = SR ...(i)

Now, in $$\triangle{ASP}$$ and $$\triangle{BQP}$$, we have,
AP = BP ...(Given)
AS = BQ ...(Given)
$$\angle{A}$$ = $$\angle{B}$$ ...(Given)
Therefore, $$\triangle{ASP}$$ $$\displaystyle \cong$$ $$\triangle{BQP}$$ ...(By SAS rule)
Therefore, SP = BQ ...(ii)(By CPCT)

Similarly, in $$\triangle{RDS}$$ and $$\triangle{RCQ}$$, we have,
SD = CQ ...(Given)
DR = RC ...(Given)
$$\angle{C}$$ = $$\angle{D}$$ ...(Given)
Therefore, $$\triangle{RDS}$$ $$\displaystyle \cong$$ $$\triangle{RCQ}$$ ...(By SAS rule)
Therefore, SR = RQ ...(iii)(By CPCT)

Thus, from eq. (i), (ii) and (iii), we can say that,
The quadrilateral PQRS is a rhombus is proved.

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. Given: ABCD is a trapezium in which AB || CD and E is mid-point of AD and EF || AB.

In $$\triangle{ABD}$$, we have,
EP || AB and E is mid-point of AD. So, by theorem, if a line drawn through the mid-point of one side of a triangle is parallel to another side, then it bisects the third side.
Therefore, P is the midpoint of BD.
Similarly, in $$\triangle{BCD}$$,
we have, PF || CD,
Therefore by converse of mid point theorem, F is mid-point of BD.

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. Given: ABCD is a parallelogram and E, F are the mid-points of sides AB and CD, respectively.

To Prove: Line segments AF and EC trisect the diagonal BD.

Proof: Since, ABCD is a parallelogram,
AB || DC and
AB = DC ...(since, Opposite sides of a parallelogram)
Thus, AE || FC and (1/2) AB = (1/2) CD
i.e., AE = FC
Therefore, AECF is a parallelogram.
Thus, AF || EC
And hence, EQ || AP and FP || CQ.

In $$\triangle{BAP}$$, E is the mid-point of AB and EQ || AP,
So, By converse of mid-point theorem,
Q is the mid-point of BP.
i.e., BQ = PQ ...(i)

Similarly, in $$\triangle{DQC}$$, F is the mid-point of DC and FP || CQ,
So, P is the mid-point of DQ.
i.e., DP = PQ ...(ii)
From Equations (i) and (ii) , we get,
BQ = PQ = DP
Hence, it is proved that CE and AF trisect the diagonal BD.

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

Let ABCD is a quadrilateral and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA, respectively.
i.e., AS = SD, AP = BP, BQ = CQ and CR = DR.
We have to show that: PR and SQ bisect each other i.e., SO = OQ and PO = OR. Now, in $$\triangle{ADC}$$, S and R are mid-point of AD and CD.
We know that, the line segment joining the mid-points of two sides of a triangle is parallel to the third side.
Thus, by midpoint theorem,
SR || AC and SR = (1/2) AC ...(i)

Similarly, in $$\triangle{ABC}$$, P and Q are mid-point of AB and BC.
Thus, by midpoint theorem,
PQ || AC and PQ = (1/2) AC ...(ii)
From Eq. (i) and (ii), we get,
i.e., PQ || SR and PQ = SR = (1/2) AC
Therefore, Quadrilateral PQRS is a parallelogram whose diagonals SQ are PR.
Also, we know that diagonals of a parallelogram bisect each other. So, and bisect each other.
Hence, it is proved that PR and SQ bisect each other.

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 is perpendicular to AC
iii) CM = MA = (1/2) AB

Given: $$\triangle{ABC}$$ is a right angled triangle i.e., $$\angle{C}$$ = $$90^\circ$$
and M is the mid-point of AB.
Also, DM || BC

i) In $$\triangle{ABC}$$, M is the mid-point of AB and BC || MD,
So, By converse of mid-point theorem,
D is the mid-point of AC.
ii) Since, BC || MD and CD is transversal.
Therefore, $$\angle{ADM}$$ = $$\angle{ACB}$$ ...(Corresponding angles)
But, $$\angle{C}$$ = $$90^\circ$$
Therefore, $$\angle{ADM}$$ = $$90^\circ$$
Hence, proved that MD is perpendicular to AC.

iii) Now, in $$\triangle{ADM}$$ and $$\triangle{ACM}$$, we have,
$$\angle{ADM}$$ = $$\angle{MDC}$$ = $$90^\circ$$ ...(Proved)
Therefore, $$\triangle{ADM}$$ $$\displaystyle \cong$$ $$\triangle{CDM}$$ ...(By SAS rule)