Mathematics Part ii Solutions Solutions for Class 9 Maths Chapter 5 - Quadrilaterals

Answer:

In a parallelogram, opposite angles are congruent. 
$\angle$XYZ = $\angle$XWZ = ${135}^{°}$ 
Also, WX || ZY
So, $\angle$YZW + $\angle$XWZ = 180$°$                   (Since, interior angles on the same side of the transversal are supplementary)
$$\begin{gathered} \Rightarrow \angle \mathrm{YZW}+135°=180° \\ \Rightarrow \angle \mathrm{YZW}=45° \end{gathered}$$
Now l(OY)= 5
we know that diagonals of a parallelogram bisect each other. 
So, l(WY) = $2\times \mathrm{OY}=2\times 5=10 \mathrm{cm}$

Answer:

Interior angles on the same side of the transversal are supplementary.
$\angle$A +$\angle$B = 180$°$
$\Rightarrow$${\left(3x + 12\right)}^{°}$ + ${\left(2x - 32\right)}^{°}$ = 180$°$
$$\begin{gathered} \Rightarrow 5x-20=180° \\ \Rightarrow 5x=200 \\ \Rightarrow x=40° \end{gathered}$$
Thus, $\angle$A = $3\times 40°+12=132°$
$\angle \mathrm{B}=2x-32=2\times 40°-32=48°$
Also, opposite angles of a parallelogram are congruent.
$$\begin{gathered} \angle \mathrm{A}=\angle \mathrm{C}=132° \\ \angle \mathrm{B}=\angle \mathrm{D}=48° \end{gathered}$$

Answer:

Perimeter = 150 cm
Let one of the sides be x.
Other side = $x+25$
Perimeter = Sum of all sides
$$\begin{gathered} x+x+\left(x+25\right)+\left(x+25\right)=150 \\ \Rightarrow 4x+50=150 \\ \Rightarrow 4x=100 \\ \Rightarrow x=25 \end{gathered}$$
Thus, the sides are 25 cm, 25 cm, 50 cm and 50 cm
 

Answer:

Let the two adjacent angles be $x \mathrm{and} 2x$.
Interior angles on the same side of the transversal are supplementary.
So, 
$$\begin{gathered} x+2x=180° \\ \Rightarrow 3x=180° \\ \Rightarrow x=60° \end{gathered}$$
Thus, the adjacent angles are $60° \mathrm{and} 120°$.
The opposite angles of a parallelogram are congruent.
So, the angles of the parallelogram will be $60°,120°$, $60° \mathrm{and} 120°$.

Answer:

In $△$AOB,
AO = 5 cm
OB = 12 cm
AB = 13 cm
5, 12 and 13 form a Pythagorean triplet.
Thus, $△$AOB is a right angle triangle, right angled at O. 
ABCD is a parallelogram. 
So, diagonals bisect each other. 
$\Rightarrow$AO = OC = 5 cm
Also, OB = OD = 12 cm.
Thus, in parallelogram ABCD, diagonals bisect at right angles.
Hence, ABCD is a rhombus. 

Answer:

In $\square$PQRS, 
$\angle$P = ${110}^{°}$ 
We know that opposite angles of a parallelogram are congruent.
So, $\angle$P = $\angle$R = ${110}^{°}$ 
In $\square$ABCR,
$\angle$R =$\angle$B = ${110}^{°}$        (Opposite angles of a parallelogram are congruent)
Interior angles on the same side of the transversal are supplementary.
So, 
$$\begin{gathered} \angle \mathrm{R}+\angle \mathrm{A}=180° \\ \Rightarrow 110°+\angle \mathrm{A}=180° \\ \Rightarrow \angle \mathrm{A}=180°-110° \\ \Rightarrow \angle \mathrm{A}=70° \end{gathered}$$
  And $\angle$A =$\angle$C = 110$°$. 

Answer:

$\square$ABCD is a parallelogram
So, AD || CB
$\Rightarrow$AD || BF
Given: AB = BE so, B is the midpoint of AE. 
By converse of mid-point theorem,
F is the midpoint of DE. So, DF = FE
In $△$EBF and $△$DCF,
DF = FE                         (Proved above)
DC = BE                        (Since DC = AB and AB = BE)
$\angle$FDC = $\angle$FEB             (Alternate interior angles of the parallel lines AE and CD)
Thus, $△$EBF $\cong$ $△$DCF     (SAS congruency)
Therefore, FB = FC             (CPCT)
Hence, ED bisects seg BC at F.    
 

Answer:

$\square$ABCD is a parallelogram,
AB $\cong$ CD and AB || CD
P and Q are the mid points of AB and CD respectively. 
So, AP $\cong$ CQ and AP || CQ
Thus, $\square$APCQ is also a parallelogram.  

Answer:

Let ABCD be a rectangle. 
$\angle$A = $\angle$B = $\angle$C = $\angle$D = $90°$.
For any quadrilateral to be a parallelogram, pair of opposite angles should be congruent. 
In rectangle ABCD, 
$\angle$A = $\angle$C =  $90°$
$\angle$B = $\angle$D = $90°$
Thus, rectangle ABCD is a paralleogram.

Answer:

G is the point of concurrence of the medians of $∆$DEF.
Let the point where the median divides EF into two equal parts be A. 
Thus, EA = AF.                     .....(1)
we know that the point of concurrence of the medians, divides each median in the ratio 2 : 1.
So, let DG = 2x  and GA = x
Given that DG = GH
So, GA = AH = x
Thus, point A dividess EF and GH into two equal parts. 
Hence, $\square$GEHF is a parallelogram as the diagonals EF and GH bisect each other. 
 

Answer:

ABCD is a parallelogram.
$$\begin{gathered} \Rightarrow \angle \mathrm{BAD}+\angle \mathrm{CDA}=180° \left(\mathrm{interior} \mathrm{angles} \mathrm{on} \mathrm{the} \mathrm{same} \mathrm{side} \mathrm{of} \mathrm{the} \mathrm{transversal} \mathrm{are} \mathrm{supplementary}\right) \\ \Rightarrow \frac{1}{2}\angle \mathrm{BAD}+\frac{1}{2}\angle \mathrm{CDA}=\frac{1}{2}\times 180° \\ \Rightarrow \angle \mathrm{SAD}+\angle \mathrm{SDA}=90° .....\left(1\right) \end{gathered}$$
$$\begin{gathered} \angle \mathrm{SAD}+\angle \mathrm{SDA}+\angle \mathrm{ASD}=180° \left(\mathrm{Angle} \mathrm{sum} \mathrm{property} \mathrm{in} △\mathrm{ASD}\right) \\ \Rightarrow 90°+\angle \mathrm{ASD}=180° \\ \Rightarrow \angle \mathrm{ASD}=90° \\ \Rightarrow \angle \mathrm{PSR}=90° \left(\mathrm{Vertically} \mathrm{opposite} \mathrm{angles}\right) \end{gathered}$$
Similarly, $\angle \mathrm{PQR}=90°$
$$\begin{gathered} \Rightarrow \angle \mathrm{DAB}+\angle \mathrm{CBA}=180° \left(\mathrm{interior} \mathrm{angles} \mathrm{on} \mathrm{the} \mathrm{same} \mathrm{side} \mathrm{of} \mathrm{the} \mathrm{transversal} \mathrm{are} \mathrm{supplementary}\right) \\ \Rightarrow \frac{1}{2}\angle \mathrm{DAB}+\frac{1}{2}\angle \mathrm{CBA}=\frac{1}{2}\times 180° \\ \Rightarrow \angle \mathrm{BAR}+\angle \mathrm{RBA}=90° \\ \mathrm{In} △\mathrm{ABR}, \\ \angle \mathrm{BAR}+\angle \mathrm{RBA}+\angle \mathrm{ARB}=180° \\ \Rightarrow 90°+\angle \mathrm{ARB}=180° \\ \Rightarrow \angle \mathrm{ARB}=90° \\ Similarly, \\ \angle \mathrm{DPC}=90° \end{gathered}$$
Thus, PQRS is a rectangle. 

Answer:

ABCD is a parallelogram
So, opposite pair of sides will be congruent and parallel.
$\Rightarrow \mathrm{AD}\cong \mathrm{BC} \mathrm{and} \mathrm{AB}\cong \mathrm{CD}$
Given, AP = BQ = CR = DS
$$\begin{gathered} \mathrm{AD}-\mathrm{SD}=\mathrm{BC}-\mathrm{BQ} \\ \Rightarrow \mathrm{AS}=\mathrm{CQ} .....\left(1\right) \end{gathered}$$
In $△$APS and $△$RCQ
AS = CQ                         (From (1))
AP = CR                          (Given)
$$\begin{gathered} \angle \mathrm{PAS}=\angle \mathrm{RCQ} \left(\mathrm{Opposite} \mathrm{angles} \mathrm{of} \mathrm{a} \mathrm{parallelogram} \mathrm{are} \mathrm{equal}\right) \\ \mathrm{Thus}, △\mathrm{APS}\cong △\mathrm{RCQ} \left(\mathrm{SAS} \mathrm{congruency}\right) \\ \Rightarrow \mathrm{SP}=\mathrm{RQ} \left(\mathrm{CPCT}\right) \end{gathered}$$
Similarly, PQ = SR
Thus, opposite pair of sides are congruent.
Hence, PQRS is a parallelogram.
 

Answer:

Diagonals of a rectangle are congruent. 
So, AC = BD = 8 cm
Also, rectangle is a parallelogram so, the diagonals bisect each other.
Thus, BO = OD = 4 cm
Now, AD || CB
So, $\angle$CAD = $\angle$ACB = ${35}^{°}$     (Since, alternate interior angles are equal)

Answer:

All the sides of a rhombus are equal. 
So, PQ = QR = 7.5 cm
A rhombus is also a parallelogram so,
$\angle$QPS = $\angle$SRQ = ${75}^{°}$                      (opposite angles of a parallelogram are congruent)
Also, $\angle$QPS + $\angle$PQR = $180°$           (interior angles on the same side of the transversal are supplementary)
$$\begin{gathered} \Rightarrow 75°+\angle \mathrm{PQR}=180° \\ \Rightarrow \angle \mathrm{PQR}=105° \end{gathered}$$ 

Answer:

Diagonals of a square are perpendicular bisectors of each other.
So, $\angle$IMJ = 90$°$.
A square has all the 4 angles as right angles. 
Also, diagonals of a square bisect the opposite angles.
So, $\angle$JIK = $\frac{1}{2}\times 90°=45°$
Similarly, $\angle$LJK = $\frac{1}{2}\times 90°=45°$
 

Answer:

Let ABCD be the rhombus.
AC = 20 cm and BD = 21 cm
We know that the diagonals of a rhombus bisect at right angles.
So, AO = OC = 10 cm
And BO = OD = 10.5 cm
In $△$AOB,
$$\begin{gathered} {\mathrm{AO}}^2+{\mathrm{OB}}^2={\mathrm{AB}}^2 \\ \Rightarrow {10}^2+10.5^2={\mathrm{AB}}^2 \\ \Rightarrow 100+110.25={\mathrm{AB}}^2 \\ \Rightarrow {\mathrm{AB}}^2=210.25 \\ \Rightarrow \mathrm{AB}=14.5 \mathrm{cm} \end{gathered}$$
Thus, each side of rhombus = 14.5 cm
Perimeter = AB + BC + CD + DA = 14.5 + 14.5 + 14.5 + 14.5 = 58 cm.
 

Answer:

Answer:

Given: side IJ || side KL
So, the interior angles on the same side of the transversal will be supplementary.
$\angle$I + $\angle$L = 180º
$\Rightarrow$${108}^{°}$ + $\angle$L = 180º
$\Rightarrow$$\angle$L = 72º
Similarly, $\angle$K + $\angle$J = 180º
$\Rightarrow$${53}^{°}$ + $\angle$J = 180º
$\Rightarrow$$\angle$J = 127º

Answer:

side  BC || side AD
Interior angles on the same side of the transversal are supplementary.
$\angle$A + $\angle$B = 180º
$\Rightarrow$${72}^{°}$ + $\angle$B = 180º
$\Rightarrow$$\angle$B = 180º − ${72}^{°}$
$\Rightarrow$$\angle$B = 108º
Construction: Draw BP || CD
So, BC || AD and BP || CD
$\Rightarrow$PBCD is a parallelogram
$\Rightarrow$CD $\cong$ BP        
Now CD = BP and CD = AB
$\Rightarrow$BP = AB
$\Rightarrow$$\angle \mathrm{BAP}=\angle \mathrm{BPA}=72°$
BP || CD so,
$\angle \mathrm{CDP}=\angle \mathrm{BPA}=72° \left(\mathrm{corresponding} \mathrm{angles} \mathrm{are} \mathrm{equal}\right)$
$\angle$B = 108º 
$\angle$D = 72º

Answer:

Construction: Draw a line CE || AB
CE || AB          (By construction)
AE || BC          (Give)
So, ABCE is a parallelogram.
Since, AB || CE
⇒ ∠BAE = ∠CED = ∠x              (Corresponding angles)         .....(1)
Now, BA $\cong$ CE                           (ABCE is a parallelogram)
and BA $\cong$ CD                              (Given)
So, CE $\cong$ CD
⇒ ∠CED = ∠CDE = ∠x               (Isosceles triangle)               .....(2)
From (1) and (2) we have
∠BAE = ∠CDE = ∠x                                                                .....(3)
∠CEA = 180º − ∠CED = 180º − ∠x
In a parallelogram, the opposite angles are equal.
So, ∠ABC = ∠CEA = 180º − ∠x                                              .....(4)
Similarly, ∠BAE = ∠BCE = ∠x 
In ∆CED,
$$\begin{gathered} \angle \mathrm{ECD}+\angle \mathrm{CED}+\angle \mathrm{EDC}=180° \\ \Rightarrow \angle \mathrm{ECD}+x+x=180º \\ \Rightarrow \angle \mathrm{ECD}=180º-2x \end{gathered}$$
Now ∠DCB = ∠BCE + ∠ECD = x + 180º − 2x = 180º − x 
Thus, $\angle$ABC$\cong$ $\angle$DCB  

Answer:

Given that X, Y and Z are the midpoints of the side AB, BC and CA respectively. 
By Midpoint theorem,
XZ || BC and XZ = $\frac{1}{2}$BC
BC = 11 cm
$\Rightarrow$XZ = $\frac{\mathrm{BC}}{2}=\frac{11}{2}=5.5 \mathrm{cm}$
Similarly, 
$$\begin{gathered} \mathrm{XY}=\frac{\mathrm{AC}}{2}=\frac{9}{2}=4.5 \mathrm{cm} \\ \mathrm{YZ}=\frac{\mathrm{AB}}{2}=\frac{5}{2}=2.5 \mathrm{cm} \end{gathered}$$

Answer:

(i) Given that M is the midpoint of PR.                                                                                   .....(1)
$\angle$PSR = $\angle$MLR = 90$°$                             (Since $\square$ PQRS and $\square$MNRL are rectangles)
Thus, LM || SP                                           (Corresponding angles are equal)                         .....(2)
From (1) and (2) we have 
L is the midpoint of SR                              (Converse of midpoint theorem)
Thus, SL = LR

(ii)                                                                         
$\angle$RNM = $\angle$RQP = 90$°$                             (Since $\square$ PQRS and $\square$MNRL are rectangles)
Thus, MN || PQ                                           (Corresponding angles are equal)                         .....(4)
From (1) and (4) we have 
N is the midpoint of RQ                              (Converse of midpoint theorem)
Join LN and SQ

By midpoint theorem, 
LN || SQ and LN = $\frac{1}{2}$SQ
 

Answer:

$∆$ABC is an equilateral triangle.
So, AB = BC = CA
$$\begin{gathered} \Rightarrow \frac{1}{2}\mathrm{AB}=\frac{1}{2}\mathrm{BC}=\frac{1}{2}\mathrm{CA} \\ \Rightarrow \mathrm{FB}=\mathrm{BD}=\mathrm{EC} .....\left(1\right) \end{gathered}$$
Points F,D and E are midpoints of side AB, side BC, side AC respectively.
By midpoint theorem,
$$\begin{gathered} \mathrm{FE}=\frac{1}{2}\mathrm{BC}=\mathrm{BD} \\ \mathrm{DE}=\frac{1}{2}\mathrm{AB}=\mathrm{FB} \\ \mathrm{FD}=\frac{1}{2}\mathrm{AC}=\mathrm{EC} \end{gathered}$$
From (1) we have FB = BD = EC
So, FE = DE = FD
Thus, $∆$FED is an equilateral traingle.

Answer:

PD is the median of QR. 
So, D is the midpoint of QR.
DN is drawn parallel to QM.
By converse of midpoint theorem, N is the midpoint of MR.            .....(1)
Similarly, T is the midpoint of PD 
Also, DN || QM
So, By converse of  midpoint theorem,
M is the midpoint of PN.                                                                    .....(2)
From (1) and (2) we have
PM = MN = NR
$$\begin{gathered} \Rightarrow \frac{\mathrm{PM}}{\mathrm{PR}}=\frac{\mathrm{PM}}{\mathrm{PM}+\mathrm{MN}+\mathrm{NR}}=\frac{\mathrm{PM}}{\mathrm{PM}+\mathrm{PM}+\mathrm{PM}}=\frac{1}{3\mathrm{PM}}=\frac{1}{3} \\ \Rightarrow \frac{\mathrm{PM}}{\mathrm{PR}}=\frac{1}{3} \end{gathered}$$
Hence Proved. 
 

Answer:

(i) When all the pairs of the adjacent sides of a quadrilateral are congruent then it is a rhombus.
Hence, the correct answer is option D. 

(ii) 

Let the diagonal be AC = 12$\sqrt{2}$ cm
since it is a squarem so, all the angles are 90$°$.
In $△$ABC, we apply the Pythagoras theorem,
$$\begin{gathered} {\mathrm{AB}}^2+{\mathrm{BC}}^2={\mathrm{AC}}^2 \\ \Rightarrow {\mathrm{AB}}^2+{\mathrm{AB}}^2={\mathrm{AC}}^2 \left(\because \mathrm{ABCD} \mathrm{is} \mathrm{a} \mathrm{square}\right) \\ \Rightarrow 2{\mathrm{AB}}^2={\mathrm{AC}}^2 \\ \Rightarrow 2{\mathrm{AB}}^2={\left(12\sqrt{2}\right)}^2 \\ \Rightarrow 2{\mathrm{AB}}^2=288 \\ \Rightarrow {\mathrm{AB}}^2=144 \\ \Rightarrow \mathrm{AB}=12 \mathrm{cm} \end{gathered}$$
So, AB = BC = CD = AD = 12 cm
Perimeter of the square ABCD = 4AB = $4\times 12 \mathrm{cm}=48 \mathrm{cm}$
Hence, the correct answer is option C. 

(iii) A rhombus is also a parallelogram so, the opposite angles will be congruent.
Thus, (2x)^° = (3x $-$ 40)°
$$\begin{gathered} \Rightarrow 3x-2x=40° \\ \Rightarrow x=40° \end{gathered}$$
Hence, the correct answer is option D. 
 

Answer:

In a rectangle, all the angles are equal to 90º. 
On applying Pythagoras theorem in $△$ABD, 
$$\begin{gathered} {\mathrm{AB}}^2+{\mathrm{AD}}^2={\mathrm{DB}}^2 \\ \Rightarrow {24}^2+7^2={\mathrm{DB}}^2 \\ \Rightarrow 576+49={\mathrm{DB}}^2 \\ \Rightarrow {\mathrm{DB}}^2=625 \\ \Rightarrow \mathrm{DB}=25 \mathrm{cm} \end{gathered}$$
Thus, the length of the diagonal is 25 cm.

Answer:

Let PQRS be the square. 
Let the diagonal SQ = 13 cm
All the angles of a square are right angles. 
In $△$SPQ, applying Pythagoras theorem,
$$\begin{gathered} {\mathrm{SP}}^2+{\mathrm{PQ}}^2={\mathrm{SQ}}^2 \\ \Rightarrow {\mathrm{SP}}^2+{\mathrm{SP}}^2={\mathrm{SQ}}^2 \left(\because \mathrm{All} \mathrm{the} \mathrm{sides} \mathrm{of} \mathrm{a} \mathrm{square} \mathrm{are} \mathrm{equal}\right) \\ \Rightarrow 2{\mathrm{SP}}^2={\mathrm{SQ}}^2 \\ \Rightarrow 2{\mathrm{SP}}^2={13}^2 \\ \Rightarrow {\mathrm{SP}}^2=\frac{169}{2} \\ \Rightarrow \mathrm{SP}=6.5\sqrt{2} \mathrm{cm} \end{gathered}$$
Thus, each side = $6.5\sqrt{2}$ cm. 

Answer:

Let the two sides be 3x and 4x . 
In a parallelogram the opposite sides are congruent.
So, the sides are 3x, 3x, 4x and 4x
Perimeter = 112 cm 
$$\begin{gathered} 3x+3x+4x+4x=112 \\ \Rightarrow 14x=112 \\ \Rightarrow x=\frac{112}{14} \\ \Rightarrow x=8 \end{gathered}$$
So,  
$$\begin{gathered} 3x=3\times 8=24 \mathrm{cm} \\ 4\mathrm{x}=4\times 8=32 \mathrm{cm} \end{gathered}$$
Thus, the sides are 24 cm, 32 cm, 24 cm and 32 cm. 

Answer:

Diagonals of a rhombus bisect each other at right angles. 
PR = 20 cm
$\Rightarrow \mathrm{PO}=\frac{1}{2}\mathrm{PR}=\frac{1}{2}\times 20=10 \mathrm{cm}$
Similarly, QS = 48 cm
$\Rightarrow \mathrm{QO}=\frac{1}{2}\mathrm{SQ}=\frac{1}{2}\times 48=24 \mathrm{cm}$
In $△$POQ, applying Pythagoras theorem,
$$\begin{gathered} {\mathrm{OQ}}^2+{\mathrm{OP}}^2={\mathrm{PQ}}^2 \\ \Rightarrow {24}^2+{10}^2={\mathrm{PQ}}^2 \\ \Rightarrow {\mathrm{PQ}}^2=576+100 \\ \Rightarrow {\mathrm{PQ}}^2=676 \\ \Rightarrow \mathrm{PQ}=26 \mathrm{cm} \end{gathered}$$

Answer:

In the given rectangle PQRS, 
$\angle$QMR = $\angle$PMS = ${50}^{°}$                        (vertically opposite angles are equal)
Also, SQ = PR                                      (Diagonals of a rectangle are equal)
So, 
$$\begin{gathered} \frac{1}{2}\mathrm{SQ}=\frac{1}{2}\mathrm{PR} \\ \Rightarrow \mathrm{SM}=\mathrm{PM} \end{gathered}$$
$\Rightarrow \angle \mathrm{MSP}=\angle \mathrm{MPS}$                                (Angle opposite to equal sides are equal)
In $△$PMS,
$$\begin{gathered} \angle \mathrm{PMS}+\angle \mathrm{MSP}+\angle \mathrm{MPS}=180° \\ \Rightarrow 50°+2\angle \mathrm{MPS}=180° \\ \Rightarrow 2\angle \mathrm{MPS}=130° \\ \Rightarrow \angle \mathrm{MPS}=65° \end{gathered}$$


 

Answer:

Given: seg AB || seg PQ , seg AB $\cong$seg PQ
So, ABQP is a parallelogram as the pair of opposite sides are congruent and parallel.
Thus, seg AP || seg BQ , seg AP $\cong$seg BQ                       .....(1)
Similarly, seg AC || seg PR, seg AC $\cong$ seg PR
So, APRC is a parallelogram. 
Thus, seg AP || seg CR, seg AP $\cong$ seg CR                       .....(2)
From (1) and (2) we have 
seg BQ || seg CR
Also, seg BQ $\cong$ seg CR
Thus, BQRC is a parallelogram as the pair of opposite sides are congruent and parallel.
Therefore, seg BC || seg QR and seg BC$\cong$seg QR as BQRC is a parallelogram. 

Answer:

Construction: Join PB and extend it to meet CD produced at R. 
To prove: PQ || AB and  PQ = $\frac{1}{2}$( AB + DC)
Proof: In $△$ABP and $△$DRP,
$$\begin{gathered} \angle \mathrm{APB}=\angle \mathrm{DPR} \left(\mathrm{Vertically} \mathrm{opposite} \mathrm{angles}\right) \\ \angle \mathrm{PDA}=\angle \mathrm{PAB} \left(\mathrm{Alternate} \mathrm{interior} \mathrm{angles} \mathrm{are} \mathrm{equal}\right) \\ \mathrm{AP}=\mathrm{PD} \left(\mathrm{P} \mathrm{is} \mathrm{the} \mathrm{mid} \mathrm{point} \mathrm{of} \mathrm{AD}\right) \end{gathered}$$
Thus, by ASA congruency, $△$ABP $\cong$ $△$DRP.
By CPCT, PB = PR and AB = RD.
In $△$BRC,
Q is the mid point of BC                     (Given)
P is the mid point of BR                      (As PB = PR)
So, by midpoint theorem, PQ || RC 
$\Rightarrow$PQ || DC
But AB || DC                                       (Given)
So, PQ || AB.
Also, PQ = $\frac{1}{2}\left(\mathrm{RC}\right)$
$$\begin{gathered} \mathrm{PQ}=\frac{1}{2}\left(\mathrm{RD}+\mathrm{DC}\right) \\ \mathrm{PQ}=\frac{1}{2}\left(\mathrm{AB}+\mathrm{DC}\right) \left(\because \mathrm{AB}=\mathrm{RD}\right) \end{gathered}$$ 


 

Answer:

Construction: Join MN. Also, join DM and produce it to AB in P. 
To prove: MN || AB
Proof: AB || DC and AC is the transversal. 
So, $\angle 1=\angle 2 \left(\mathrm{Alternate} \mathrm{angles} \mathrm{are} \mathrm{equal}\right)$
In $△$AMP and $△$CMD,
$\angle 1=\angle 2 \left(\mathrm{Alternate} \mathrm{angles} \mathrm{are} \mathrm{equal}\right)$
$\angle 3=\angle 4$(Vertically opposite angles)
AM = MC             (M is midpoint of AC)
So, $△$AMP $\cong$ $△$CMD          (ASA congruency criteria)
By CPCT, DM = MP.
So, M is the midpoint of DP.
In $△$DPB, 
M is the midpoint of DP and N is the midpoint of DB.
By, midpoint theorem MN || AB.