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

Given : A quadrilateral ABCD in which혻 P, Q, R, and혻 S are혻 respectively the mid- points of the sides AB, BC, CD and DA. Also혻 AC is its diagonal.
To prove :
(i) SR || AC and혻 $SR = {1 \over 2}AC$
(ii)혻 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 혻혻 혻$SR||\,AC\,and\,SR = {1 \over 2}AC$혻혻 혻혻혻 혻[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, 혻혻혻 PQ || AC
and, 혻혻 혻혻혻혻 $PQ = {1 \over 2}AC$ 혻 혻 혻 혻 혻 혻 혻 혻 혻 혻 혻 혻 혻 혻 혻 혻 혻 혻 혻 혻 혻[Mid- point theorem]
Thus,혻 we have proved that :
$\left. {\matrix{{PQ||AC} \cr {SR||AC} \cr} } \right\} \Rightarrow PQ||SR$
Also 혻혻$\left. {\matrix{{PQ = {1 \over 2}AC} \cr{SR = {1 \over 2}AC} \cr} } \right\} \Rightarrow PQ||SR$

(iii) Since PQ = SR and PQ|| SR
$\Rightarrow$혻 One pair혻 of opposite혻 sides are equal혻 and parallel.
$\Rightarrow$혻혻 PQRS is a parallelogram.

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

Given : ABCD is rhombus in which혻 P, Q, R and혻 S are the mid points of AB, BC, CD and DA respectively. PQ,혻 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 혻혻혻 혻PQ || AC and $PQ = {1 \over 2}AC$혻혻혻혻혻혻혻혻혻혻혻 [Mid-point Theorem]
Similarly, in $\Delta$ ADC , R and S are혻 the mid- points of CD and AD.
Therefore 혻혻혻 혻SR || AC and $SR = {1 \over 2}AC$혻혻혻혻혻혻혻혻혻혻혻혻 [Mid-point Theorem]
From (1) and (2) , we get
PQ || RS and PQ = SR
Now , in quadrilateral PQRS its one pair혻 of opposite혻 sides PQ and SR is equal혻 and parallel.
Therefore혻 PQRS is혻 a parallelogram
Therefore 혻혻혻 혻AB = BC 혻혻 혻혻혻 혻혻혻 혻혻혻 혻[Sides of a rhombus]
$\Rightarrow$혻혻 혻${1 \over 2}AB = {1 \over 2}BC$
$\Rightarrow \,PB = BQ$

$\Rightarrow$혻혻 혻$\angle 3 = \angle 4$혻혻 혻혻혻 혻[$\angle s$ opp. to equal sides of혻 a triangle]
Now혻 in $\Delta$ APS and $\Delta$ CQR we have
AP = CQ 혻혻 혻혻혻 혻혻혻 혻혻혻 혻[Halves of equal sides AB, BC]
AS = CR혻혻 혻혻혻 혻혻혻 혻혻혻 혻[Halves of equal sides AD, CD]
PS = QR 혻혻 혻혻혻 혻혻혻 혻혻혻 혻[Opp. sides of parallelogram PQRS]
Therefore 혻혻 혻$\Delta APS \cong \Delta CQR$혻혻 혻[SSS Cong. Theorem]
$\Rightarrow$혻혻 혻$\angle 1 = \angle 2$혻혻 혻혻혻 혻[Corresponding parts of congruent triangles are equal]
Now , $\angle 1 + \angle SPQ + \angle 3 = 180^o$혻혻 혻[Linear pair혻 axiom]
Therefore $\angle 1 + \angle SPQ + \angle 3 = \angle 2 + \angle PQR + \angle 4$
But 혻혻혻 혻혻혻 혻$\angle 1 = \angle 2\,\,and\,\,\angle 3 = \angle 4$혻혻 혻[Proved above]
Therefore 혻혻 혻혻혻 혻$\angle SPQ = \angle PQR$
Since혻혻 혻SP|| RQ , and PQ혻 intersects them ,
Therefore 혻혻 혻$\angle SPQ + \angle PQR = 180^o$혻혻혻혻혻 [Since consecutive interior angles are supplementary]
From혻 (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 angle혻 $\angle SPQ = 90^o$.
Hence PQRS is a rectangle.

Q.3혻혻혻 ABCD is a rectangle혻 and P, Q, R and S are혻 mid- points of the sides AB, BC, CD and DA respectively. Show혻 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, 혻혻혻혻 PQ || AC and $PQ = {1 \over 2}AC$ 혻 [Mid-point theorem]혻 혻혻혻 혻혻혻혻 ... (1)
Similarly, in $\Delta$ ADC, R and S are the mid- points of sides CD and AD.
Therefore 혻혻혻 혻SR|| AC and혻 $SR = {1 \over 2}AC$혻혻 혻혻[Mid-point theorem] 혻 혻혻혻 혻... (2)
From혻 (1) and (2), we get
PQ || SR and PQ = SR 혻 혻 혻 혻 혻 혻 혻 혻 혻 혻 혻 혻 혻 혻 혻 혻 혻 혻 혻 혻 혻 혻 혻 혻 혻 혻 혻 혻 혻 혻 혻 혻 혻 혻 혻 혻 혻 혻... (3)
Now혻 in quad. PQRS, its one pair혻 of opposite혻 side PQ and SR is parallel and equal. 혻혻 혻혻혻 혻[From (3)]
Therefore PQRS is a parallelogram 혻 혻 혻 혻 혻 혻 혻 혻 혻 혻 혻 혻 혻 혻 혻 혻 혻 혻 혻 혻 혻 혻 혻 ... (4)
Now, 혻혻 혻AD = BC 혻혻 혻혻혻 혻혻혻 혻혻혻 혻[Opp. sides of rect. ABCD]
$\Rightarrow$혻혻 혻${1 \over 2}AD = {1 \over 2}BC\,\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,AS = BQ$
In $\Delta$ APS and BPQ , we have 혻혻 혻
AP = BP 혻 혻 혻 혻 혻 혻 혻 혻 혻 혻 혻 혻 혻 혻 혻 혻 혻 혻 혻 혻 혻 혻 혻 혻 혻 [Since P is the mid- point of AB]
$\angle PAS = \angle PBQ$ 혻 혻 혻 혻 혻 혻 [Each = 90쨘]
AS = BQ 혻 혻 혻 혻 혻 혻 혻 혻 혻 혻 혻 혻 혻 혻 혻 혻 혻 혻 혻 혻 혻 혻 혻 혻 혻 혻 혻 혻 혻 [Proved above]
Therefore 혻혻 혻$\Delta APS \cong \Delta BPQ$ 혻혻혻 혻[SAS congruence axiom]
$\Rightarrow$혻혻 혻PS = PQ 혻 혻 혻 혻 혻 혻 혻 혻 [Corresponding parts of congruent triangles are equal] 혻 혻 혻혻 ... (5)
From혻 (4) and (5) we get,
PQ = QR = RS = PS
PQRS is a rhombus.

Q.4혻혻혻혻 ABCD is a trapezium in which혻 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). Show혻 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 혻혻 혻혻혻 혻[Given]
EG || AB혻혻 혻혻혻 혻혻혻 혻혻혻 혻혻혻 혻[Since EF || AB] Therefore by converse of mid- point theorem G is혻 the mid- point of DB.
In $\Delta$ BCD, G is the혻 mid- point of BD혻혻 혻혻 혻 혻 혻 혻 혻 혻 혻 혻 혻 [Proved]
GF || DC혻혻 혻혻혻 혻[Since AB|| DC, EF || AB $\Rightarrow$ DC|| EF]
Therefore혻 by converse of mid- point theorem -
F is the mid- point of BC.

Q.5혻혻혻 In a parallelogram ABCD, E and F are the mid- points of sides AB and CD respectively (see figure). Show혻 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 혻 혻 혻 혻 혻 혻 혻 혻 혻 혻 [Given]
Therefore 혻혻 혻혻혻 혻AB || DC and AB = DC혻혻 혻혻혻 혻[Opp. sides of parallelogram]
E is the mid- point of AB 혻 혻 혻 혻 혻 혻 혻 혻 혻 혻 혻 혻 혻 혻 혻 혻 [Given]
Therefore 혻혻 혻혻혻 혻$AE = {1 \over 2}AB$혻혻 혻혻혻 혻혻혻 혻혻혻 혻... (1)
F is the mid- point of CD
Therefore 혻혻혻 혻혻혻 혻$CF = {1 \over 2}CD$
$\Rightarrow$혻혻 혻혻혻 혻$CF = {1 \over 2}AB$혻혻 혻혻혻 혻[Since CD = AB]혻혻 혻... (2)
From (1) and (2) ,
AE = CF
Also 혻혻 혻혻혻 혻AE || CF 혻 혻 혻 혻 혻 혻 혻 혻 혻 혻 혻 혻 혻 혻 혻 혻 혻 혻 혻 혻 혻 혻 혻 혻 혻 혻 혻 혻 혻 혻 혻 혻 혻 혻 혻 혻 혻[Since AB || DC]
Thus, a pair혻 of opposite혻 sides of a quadrilateral혻 AECF are parallel and equal.
Quadrilateral AECF is a parallelogram .
$\Rightarrow$혻혻 혻혻혻 혻EC || AF
$\Rightarrow$혻혻 혻혻혻 혻EQ|| AP and QC || PF
In $\Delta$ BPA , E is the mid- point혻 of BA 혻혻 혻혻혻 혻[Given]
EQ|| AP 혻 혻 혻 혻 혻 혻 혻 혻 혻 혻 혻 혻 혻 혻[Proved]
Therefore, 혻혻 혻혻혻혻 BQ = PQ혻혻 혻혻혻 혻[Converse of mid- point theorem]혻혻 혻... (3)
Similarly by taking혻 $\Delta$ CQD,혻 we can prove that
$\Rightarrow$혻혻DP = QP
From (3) and (4), we get
$\Rightarrow$혻혻혻혻 BQ = QP = PD
Hence , AF and CE trisect the diagonal BD.

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

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

Proof : In $\Delta$ ABC, P and Q are mid- points of AB and BC respectively.
Therefore 혻혻 혻PQ|| AC and $PQ = {1 \over 2}AC$혻혻혻혻혻혻혻혻혻 [Mid-point theorem].........(1)
Similarly, we can prove that
RS || AC and $RS = {1 \over 2}AC$혻혻혻혻 [Mid-point theorem]...................(2)
From (1) and (2)-
Therefore 혻혻혻 혻PQ || SR and PQ = SR
Thus, a pair혻 of opposite혻 sides of a quadrilateral PQRS are parallel and equal.
Therefore , quadrilateral혻 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.,혻 the line segements joining혻 the mid- points of opposite 혻sides of quadrilateral혻 ABCD bisect each other.

Q.7혻혻혻혻 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. Show혻 that
혻혻혻혻혻혻혻혻혻혻혻
(i) D is the mid-point of AC
혻혻혻혻혻혻혻혻혻혻혻
(ii) MD $\bot$ AC
혻혻혻혻혻혻혻혻혻혻혻
(iii) $CM = MA = {1 \over 2}AB$
Sol.

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

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

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

(iii) In $\Delta s$ AMD and CMD, we have
AD = CD 혻혻 혻혻혻 혻혻혻 혻혻혻 혻[From (1)]
$\angle ADM = \angle CDM$혻혻 혻혻혻 혻[From (2)]
and, 혻혻 혻혻혻 혻 MD = MD 혻혻 혻혻혻 혻혻혻 혻[Common]
Therefore,혻 by혻 SAS criterion혻 of congruence
$\Delta \,AMD\, \cong \,\Delta CMD$
$\Rightarrow$혻혻 혻혻혻 혻MA혻 = MC 혻 혻 혻 혻혻 [Since corresponding혻 parts of congruent triangles are equal]
혻Also, 혻혻 혻$MA = {1 \over 2}AB$, since M is the mid-point of AB
Hence, 혻혻 혻$CM = MA = {1 \over 2}AB$

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