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.
Answer :

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,
angles of quadrilateral
=> 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.
Answer :

Given: Parallelogram is ABCD whose diagonals AC and BD are equal.
image
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.
Answer :

Given: A quadrilateral ABCD whose diagonals AC and BD bisect each other at right angles.
i.e., OA = OC and OB = OD
Also, \(\angle{AOD}\) = \(\angle{AOB}\) = \(\angle{COD}\) = \(\angle{BOC}\) = \(90^\circ\).
image
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.
Answer :

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

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

Proof: In \(\triangle{ABC}\) and \(\triangle{BAD}\), we have,
BC = AD ...(given)
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.
Answer :

Given: A quadrilateral ABCD in which AC = BD and AC \(\perp{BD}\) such that OA = OC and OB = OD. So, ABCD is a parallelogram.
image
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)
BC = AD ...(from(i))
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.
image
Answer :

Given, diagonal AC of a parallelogram ABCD bisects \(\angle{A}\).
image
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 \(\angle{D}\).
Answer :

Given: ABCD is a rhombus. i.e., AD = AB = BC = CD ... (i)
image
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,
AD = AB ...(Given)
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)
AD = CD ...(Given)
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 \(\angle{D}\).
Answer :

Given: ABCD is a rectangle. AB = CD and BC = AD ...(i)
image

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,
AD = CD ...(Given)
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.
image
Answer :

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
image
Answer :

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).
image
Show that
(i) Quadrilateral ABED is a parallelogram
(ii) Quadrilateral BEFC is a parallelogram
(iii) AD || CF and AD = CF
(iv) Quadrilateral ACFD is a Parallelogram
(v) AC = DF
(vi) \(\triangle{ABC}\) \(\displaystyle \cong \) \(\triangle{DEF}\)
Answer :

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

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

(ii) In quadrilateral BEFC,
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,
AD || BE and AD = BE ...(i)
Also, BEFC in a parallelogram,
CF || BE and CF = BE ...(ii)
Thus, from Eq. (i) and (ii), we get,
AD || CF and AD = CF ...(From part (iii))
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).
image
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].
Answer :

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.
image
Now, ADCE is a parallelogram.
Therefore, AD || CE and AD = CE
But AD = BC
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,
AD = BC ...(Given)
\(\angle{A}\) = \(\angle{B}\) ...(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.
image
Answer :

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.
Answer :

image
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.
image
Answer :

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.
image
Answer :

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.
image
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.
image
Answer :

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.
Answer :

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.
image
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
Answer :

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.
i.e., AD = CD ...(i)
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,
AD = CD ...(from (i))
DM = MD ...(Common side)
\(\angle{ADM}\) = \(\angle{MDC}\) = \(90^\circ\) ...(Proved)
Therefore, \(\triangle{ADM}\) \(\displaystyle \cong \) \(\triangle{CDM}\) ...(By SAS rule)
Therefore, CM = AM ...(ii)(By CPCT)

Also, M is the mid point of AB.
i.e., AM = BM = (1/2) AB ...(iii)
Thus, from (ii) and (iii),
CM = AM = (1/2) AB
Hence, proved.