# Quadrilaterals : Exercise 8.2 (Mathematics NCERT Class 9th) Q.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 $SR = {1 \over 2}AC$
(ii) PQ = SR
(iii) PQRS is a parallelogram.

Given : A quadrilateral ABCD in which  P, Q, R, and  S are  respectively the mid- points of the sides AB, BC, CD and DA. Also  AC is its diagonal.
To prove :
(i) SR || AC and  $SR = {1 \over 2}AC$
(ii)  PQ = SR
(iii) PQRS is a parallelogram.

Proof : (i) In $\Delta ACD$, we have S is the mid- point of AD and R is the mid- point of CD.
Then     $SR||\,AC\,and\,SR = {1 \over 2}AC$        [Mid- point theorem]

(ii) In $\Delta$ ABC, we have P is the mid- point of the side AB and Q is the mid- point the side BC.
Then,     PQ || AC
and,         $PQ = {1 \over 2}AC$                                          [Mid- point theorem]
Thus,  we have proved that :
$\left. {\matrix{{PQ||AC} \cr {SR||AC} \cr} } \right\} \Rightarrow PQ||SR$
Also   $\left. {\matrix{{PQ = {1 \over 2}AC} \cr{SR = {1 \over 2}AC} \cr} } \right\} \Rightarrow PQ||SR$

(iii) Since PQ = SR and PQ|| SR
$\Rightarrow$  One pair  of opposite  sides are equal  and parallel.
$\Rightarrow$   PQRS is a parallelogram.

Q.2    ABCD is a rhombus and P, Q, R and S are respectively the mid- points of the sides AB, BC, CD and DA respectively. Show  that the quadrilateral  PQRS is rectangle.
Sol.

Given : ABCD is rhombus in which  P, Q, R and  S are the mid points of AB, BC, CD and DA respectively. PQ,  OR, RS and SP are joined to obtain a quadrilateral PQRS.

Proof : In $\Delta$ ABC, P and Q are the mid- points of AB and BC.
Therefore      PQ || AC and $PQ = {1 \over 2}AC$            [Mid-point Theorem]
Similarly, in $\Delta$ ADC , R and S are  the mid- points of CD and AD.
Therefore      SR || AC and $SR = {1 \over 2}AC$             [Mid-point Theorem]
From (1) and (2) , we get
PQ || RS and PQ = SR
Now , in quadrilateral PQRS its one pair  of opposite  sides PQ and SR is equal  and parallel.
Therefore  PQRS is  a parallelogram
Therefore      AB = BC                 [Sides of a rhombus]
$\Rightarrow$    ${1 \over 2}AB = {1 \over 2}BC$
$\Rightarrow \,PB = BQ$

$\Rightarrow$    $\angle 3 = \angle 4$        [$\angle s$ opp. to equal sides of  a triangle]
Now  in $\Delta$ APS and $\Delta$ CQR we have
AP = CQ                 [Halves of equal sides AB, BC]
AS = CR                [Halves of equal sides AD, CD]
PS = QR                 [Opp. sides of parallelogram PQRS]
Therefore     $\Delta APS \cong \Delta CQR$    [SSS Cong. Theorem]
$\Rightarrow$    $\angle 1 = \angle 2$        [Corresponding parts of congruent triangles are equal]
Now , $\angle 1 + \angle SPQ + \angle 3 = 180^o$    [Linear pair  axiom]
Therefore $\angle 1 + \angle SPQ + \angle 3 = \angle 2 + \angle PQR + \angle 4$
But          $\angle 1 = \angle 2\,\,and\,\,\angle 3 = \angle 4$    [Proved above]
Therefore         $\angle SPQ = \angle PQR$
Since    SP|| RQ , and PQ  intersects them ,
Therefore     $\angle SPQ + \angle PQR = 180^o$      [Since consecutive interior angles are supplementary]
From  (3) and (4), we get
$\angle PQR+\angle PQR = 180^o$
2 $\angle PQR = 180^o$
$\angle PQR = 90^o$
$\angle SPQ = \angle PQR = 90^o$
Thus , PQRS is a parallelogram whose one angle  $\angle SPQ = 90^o$.
Hence PQRS is a rectangle.

Q.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.
Sol.

Given : ABCD is a rectangle in which P, Q, R, and S are the mid- points of AB, BC, CD and DA respectively.PQ, QR, RS and SP are joined to obtain a quadrilateral PQRS.

Proof : In $\Delta$ ABC, P and Q are the mid- points of sides AB and BC.
Therefore,      PQ || AC and $PQ = {1 \over 2}AC$   [Mid-point theorem]           ... (1)
Similarly, in $\Delta$ ADC, R and S are the mid- points of sides CD and AD.
Therefore      SR|| AC and  $SR = {1 \over 2}AC$     [Mid-point theorem]        ... (2)
From  (1) and (2), we get
PQ || SR and PQ = SR                                                                            ... (3)
Now  in quad. PQRS, its one pair  of opposite  side PQ and SR is parallel and equal.         [From (3)]
Therefore PQRS is a parallelogram                                               ... (4)
Now,     AD = BC                 [Opp. sides of rect. ABCD]
$\Rightarrow$    ${1 \over 2}AD = {1 \over 2}BC\,\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,AS = BQ$
In $\Delta$ APS and BPQ , we have
AP = BP                                                   [Since P is the mid- point of AB]
$\angle PAS = \angle PBQ$             [Each = 90º]
AS = BQ                                                           [Proved above]
Therefore     $\Delta APS \cong \Delta BPQ$      [SAS congruence axiom]
$\Rightarrow$    PS = PQ                 [Corresponding parts of congruent triangles are equal]        ... (5)
From  (4) and (5) we get,
PQ = QR = RS = PS
PQRS is a rhombus.

Q.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 : In trapezium ABCD, AB || DC
E is the mid- point of AD, EF || AB.
To prove : F is the mid- point of BC.
Construction : Join DB. Let it intersects EF in G.

Proof : In $\Delta$ DAB, E is the mid- point of AD         [Given]
EG || AB                    [Since EF || AB] Therefore by converse of mid- point theorem G is  the mid- point of DB.
In $\Delta$ BCD, G is the  mid- point of BD                        [Proved]
GF || DC        [Since AB|| DC, EF || AB $\Rightarrow$ DC|| EF]
Therefore  by converse of mid- point theorem -
F is the mid- point of BC.

Q.5    In a parallelogram ABCD, E and F are the mid- points of sides AB and CD respectively (see figure). Show  that the line segements AF and EC trisect the diagonal BD.

Given : E and F are the mid- points of sides AB and CD of the parallelogram ABCD whose diagonal is BD.
To prove : BQ = QP = PD

Proof : ABCD is parallelogram                     [Given]
Therefore         AB || DC and AB = DC        [Opp. sides of parallelogram]
E is the mid- point of AB                                 [Given]
Therefore         $AE = {1 \over 2}AB$                ... (1)
F is the mid- point of CD
Therefore          $CF = {1 \over 2}CD$
$\Rightarrow$        $CF = {1 \over 2}AB$        [Since CD = AB]    ... (2)
From (1) and (2) ,
AE = CF
Also         AE || CF                                                                          [Since AB || DC]
Thus, a pair  of opposite  sides of a quadrilateral  AECF are parallel and equal.
Quadrilateral AECF is a parallelogram .
$\Rightarrow$        EC || AF
$\Rightarrow$        EQ|| AP and QC || PF
In $\Delta$ BPA , E is the mid- point  of BA         [Given]
EQ|| AP                            [Proved]
Therefore,         BQ = PQ        [Converse of mid- point theorem]    ... (3)
Similarly by taking  $\Delta$ CQD,  we can prove that
$\Rightarrow$  DP = QP
From (3) and (4), we get
$\Rightarrow$     BQ = QP = PD
Hence , AF and CE trisect the diagonal BD.

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

Given : In a quadrilateral ABCD, P, Q, R and S are respectively the mid- points of AB, BC, CD and DA. PR and QS  intersect each other at O. To prove : OP  = OR,  OQ = OS.
Construction : Join PQ,  QR, RS, SP, AC and BD.

Proof : In $\Delta$ ABC, P and Q are mid- points of AB and BC respectively.
Therefore     PQ|| AC and $PQ = {1 \over 2}AC$          [Mid-point theorem].........(1)
Similarly, we can prove that
RS || AC and $RS = {1 \over 2}AC$     [Mid-point theorem]...................(2)
From (1) and (2)-
Therefore      PQ || SR and PQ = SR
Thus, a pair  of opposite  sides of a quadrilateral PQRS are parallel and equal.
Therefore , quadrilateral  PQRS is a parallelogram.
Since the diagonals of a parallelogram bisect each other.
OP = OR = and OQ = OS
Therefore , diagonals PR and QS of a || gm PQRS i.e.,  the line segements joining  the mid- points of opposite  sides of quadrilateral  ABCD bisect each other.

Q.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 $\bot$ AC

(iii) $CM = MA = {1 \over 2}AB$
Sol.

Given: $\Delta$ ABC is right angled at C, M is the mid- point  of hypotenuse AB. Also MD||BC. To prove that :
(i)  D is the mid- points of AC
(ii)  MD $\bot$ AC
(iii)  $CM = MA = {1 \over 2}AB$

Proof : (i) In $\Delta$ ABC, M is the mid- point  of AB and MD|| BC. Therefore , D is the mid- point of AC.
i.e.,         AD = DC                                                                   ... (1)

(ii) Since MD || BC, Therefore,
$\angle ADM = \angle ACB$         [Corresponding angles]
$\Rightarrow$    $\angle ADM = 90^o$            [Since $\angle ACB = 90^o$ (given)]
But, $\angle ADM + \angle CDM = 180^o$               [Since $\angle ADM\,\,and\,\,\angle CDM$ are angles of a linear pair]
Therefore,      $90^o + \angle CDM = 180^o\,\,\,\, \Rightarrow \,\,\angle CDM = 90^o$
Thus         $\angle ADM = \angle CDM = 90^o$        ... (2)
$\Rightarrow$    $MD \bot AC$

(iii) In $\Delta s$ AMD and CMD, we have
$\angle ADM = \angle CDM$        [From (2)]
$\Delta \,AMD\, \cong \,\Delta CMD$
$\Rightarrow$        MA  = MC          [Since corresponding  parts of congruent triangles are equal]
Also,     $MA = {1 \over 2}AB$, since M is the mid-point of AB
Hence,     $CM = MA = {1 \over 2}AB$