NCERT Solutions Class 9 Maths Chapter 8 Quadrilaterals

3.0NCERT Questions with Solution

Exercise: 8.1

1. If the diagonals of a parallelogram are equal, then show that it is a rectangle.
   
   Sol. Given : ABCD is a parallelogram with diagonal $\mathrm{AC}=$ diagonal BD To prove : ABCD is a rectangle. Proof : In triangle ABC and ABD, $\mathrm{AB}=\mathrm{AB}$ [Common] $\mathrm{AC}=\mathrm{BD}$ [Given] $\mathrm{AD}=\mathrm{BC}[0\mathrm{pp}$. Sides of a $\Vert \mathrm{gm}]$ $\therefore \Delta \mathrm{ABC}\cong △\mathrm{BAD}$ [By SSS congruency] $\Rightarrow \angle \mathrm{DAB}=\angle \mathrm{CBA}$ [By C.P.C.T.] $[\because AD\Vert BC$ and $AB$ cuts them, the sum of the interior angle of the same side of transversal is $180^{\circ }$ ] $\angle \mathrm{DAB}+\angle \mathrm{CBA}=180^{\circ }$ From eq. (i) and (ii), $\angle \mathrm{DAB}=\angle \mathrm{CBA}=90^{\circ }$ So, ABCD is a parallelogram with one of the angles equal to $90^{\circ }$, Hence, ABCD is a rectangle.
2. Show that the diagonals of a square are equal and bisect each other at right angles. Sol. Given: ABCD is a square. To Prove : (i) AC = BD (ii) AC and BD bisect each other at right angles.
   
   Proof: In $△ABC$ and $△BAD$, $\mathrm{AB}=\mathrm{BA}$ [Common] $\mathrm{BC}=\mathrm{AD}$ [Opp. sides of square ABCD] $\angle \mathrm{ABC}=\angle \mathrm{BAD}[$ Each $=90^{\circ }(\because \mathrm{ABCD}$ is a square $)]$ $\therefore △\mathrm{ABC}\cong △\mathrm{BAD}$ [SAS Rule] $\therefore \mathrm{AC}=\mathrm{BD}$... (i) [C.P.C.T.] In $△AOD$ and $△BOC$ $\mathrm{AD}=\mathrm{CB}$ [Opp. sides of square ABCD ] $\angle \mathrm{OAD}=\angle \mathrm{OCB}$ [Alternate angles as $\mathrm{AD}\Vert \mathrm{BC}$ and transversal AC intersects them] $\angle \mathrm{ODA}=\angle \mathrm{OBC}$ [Alternate angles as $\mathrm{AD}\Vert \mathrm{BC}$ and transversal BD intersects them] $△\mathrm{AOD}\cong △\mathrm{COB}$ [ASA Rule] $\therefore \mathrm{OA}=\mathrm{OC}$ and $\mathrm{OB}=\mathrm{OD}$ ...(ii)[C.P.C.T.] So, 0 is the midpoint of $AC$ and $BD$. Now, In $△AOB$ and $△COB$ $\begin{array}{ll}\mathrm{AB}=\mathrm{BC}& \text{ [Given] }\\ OA=OC& \text{ [from (ii)] }\\ OB=OB& [\text{ Common] }\\ \therefore △AOB\cong △COB& [\text{ By SSS Rule] }\\ \therefore \angle AOB=\angle BOC& [\text{ C.P.C.T] }\end{array}$ But $\angle \mathrm{AOB}+\angle \mathrm{BOC}=180^{\circ }$ [Linear pair $]$ $\angle \mathrm{AOB}+\angle \mathrm{AOB}=180^{\circ }$ [ $\angle \mathrm{AOB}=\angle \mathrm{BOC}$ proved earlier] $\Rightarrow 2\angle \mathrm{AOB}=180^{\circ }$ $\Rightarrow \angle \mathrm{AOB}=\frac{180^{\circ }}{2}=90^{\circ }$ $\therefore \angle \mathrm{AOB}=\angle \mathrm{BOC}=90^{\circ }$ $\therefore \mathrm{AC}$ and BD bisect each other at right angles.
3. Diagonal $AC$ of a parallelogram $ABCD$ bisects $\angle \mathrm{A}$. Show that (i) it bisects $\angle \mathrm{C}$ also (ii) ABCD is a rhombus.
   
   
4. Diagonal $AC$ bisects $\angle A$ of the parallelogram ABCD. To prove: (i) AC bisects $\angle \mathrm{C}$ (ii) ABCD is a rhombus Proof: (i) Since $\mathrm{AB}|\mid \mathrm{DC}$ and AC intersects them. $\therefore \angle 1=\angle 3$ [Alternate angles] ...(i) Similarly, $\angle 2=\angle 4$ But $\angle 1=\angle 2$ [Given] $\therefore \angle 3=\angle 4$ [Using eq. (i), (ii) and (iii)] Thus, AC bisects $\angle \mathrm{C}$. (ii) since $\angle 2=\angle 3$ (using (i) and (iii)) $\Rightarrow \mathrm{AD}=\mathrm{CD}$ [Sides opposite to equal angles] Also, ABCD is a parallelogram. $\Rightarrow AD=BC$ and $AB=CD$ $\therefore \mathrm{AB}=\mathrm{CD}=\mathrm{AD}=\mathrm{BC}$ Hence, ABCD is a rhombus.
5. ABCD is a rectangle in which diagonal AC bisects $\angle \mathrm{A}$ as well as $\angle \mathrm{C}$. Show that (i) ABCD is a square (ii) diagonal BD bisects $\angle \mathrm{B}$ as well as $\angle \mathrm{D}$. Sol. Given : ABCD is a rectangle in which diagonal AC bisects $\angle \mathrm{A}$ as well as $\angle \mathrm{C}$.
   
   To prove : (i) ABCD is a square (ii) diagonal BD bisects $\angle \mathrm{B}$ as well as $\angle \mathrm{D}$. Proof: (i) $\because \mathrm{AB}\Vert \mathrm{DC}$ and transversal AC intersects them. So, $\angle \mathrm{BAC}=\angle \mathrm{DCA}$ [ Alternate angles] But $\angle \mathrm{BAC}=\angle \mathrm{DAC}[\because \mathrm{AC}$ bisects $\angle \mathrm{A}]$ $\therefore \angle \mathrm{DCA}=\angle \mathrm{DAC}$ $\Rightarrow \mathrm{DA}=\mathrm{CD}$ [Sides opposite to equal angles of a triangle are equal] But $\mathrm{AB}=\mathrm{CD}$ and $\mathrm{DA}=\mathrm{BC}$ [Opposite side of a rectangle] $\therefore \mathrm{AB}=\mathrm{BC}=\mathrm{CD}=\mathrm{DA}$ Also, $\angle \mathrm{A}=\angle \mathrm{B}=\angle \mathrm{C}=\angle \mathrm{D}=90^{\circ }$ $[\because \mathrm{ABCD}$ is a rectangle] Hence, ABCD is a square. (ii) $\mathrm{In}△\mathrm{BAD}$ and $△\mathrm{BCD}$, $$\begin{gathered} BA=BC[\because ABCD\text{ is a square }] \\ AD=CD[\because ABCD\text{ is a square }] \\ BD=BD[Common] \\ \therefore \Delta \mathrm{BAD}\cong △\mathrm{BCD} \\ [By\text{ SSS congruence rule }] \\ \therefore \angle \mathrm{ABD}=\angle \mathrm{CBD}\text{ [By C.P.C.T.] } \\ \angle \mathrm{ADB}=\angle \mathrm{CDB}[\text{ By C.P.C.T. }] \end{gathered}$$ Hence, diagonal BD bisects $\angle \mathrm{B}$ as well as $\angle \mathrm{D}$
6. In parallelogram $ABCD$, two points $P$ and Q are taken on diagonal BD such that $\mathrm{DP}=$ BQ. Show that: (i) $△\mathrm{APD}\cong △\mathrm{CQB}$ (ii) $\mathrm{AP}=\mathrm{CQ}$ (iii) $△\mathrm{AQB}\cong △\mathrm{CPD}$ (iv) $\mathrm{AQ}=\mathrm{CP}$ (v) APCQ is a parallelogram
   
   Sol. (i) In $△APD$ and $△CQB$, we have $\mathrm{DP}=\mathrm{BQ}$ [Given] $\mathrm{AD}=\mathrm{CB}$ [Opposite sides of parallelogram ABCD] $\angle \mathrm{ADP}=\angle \mathrm{CBQ}$ [Pair of alternate angles] $\Rightarrow △\mathrm{APD}\cong △\mathrm{CQB}$ [SAS congruence criteria] (ii) Then, by CPCT, we have AP = CQ (iii) We can prove $\Delta AQB\cong △CPD$ [as we have done in (i)] (iv) By CPCT, we have $\mathrm{AQ}=\mathrm{CP}$ (v) Now, we have $\mathrm{AP}=\mathrm{CQ}$ and $\mathrm{AQ}=\mathrm{CP}$ Hence, APCQ is a parallelogram.
7. $ABCD$ is a parallelogram and $AP$ and $CQ$ are perpendiculars from vertices A and C on diagonal BD. Show that (i) $△\mathrm{APB}\cong △\mathrm{CQD}$ (ii) AP = CQ
   
   Sol. Given : ABCD is a parallelogram and AP and CQ are perpendiculars from vertices $A$ and $C$ on diagonal $BD$ respectively. To prove : (i) $△\mathrm{APB}\cong △\mathrm{CQD}$ (ii) $\mathrm{AP}=\mathrm{CQ}$ Proof: (i) In $△\mathrm{APB}$ and $△CQD$, $\mathrm{AB}=\mathrm{CD}[$ Opposite side of ॥ gm ABCD] $\angle \mathrm{ABP}=\angle \mathrm{CDQ}$ $[\because \mathrm{AB}|\mid \mathrm{DC}$ and transversal BD intersect them] $\angle \mathrm{APB}=\angle \mathrm{CQD}$ $[$ Each $=90^{\circ }]$ $\therefore △\mathrm{APB}\cong △\mathrm{CQD}$ [AAS Rule] (ii) $\mathrm{AP}=\mathrm{CQ}$ [C.P.C.T.]
8. ABCD is a trapezium in which $\mathrm{AB}\Vert \mathrm{CD}$ and $AD=BC$. Show that (figure) (i) $\angle \mathrm{A}=\angle \mathrm{B}$ (ii) $\angle \mathrm{C}=\angle \mathrm{D}$ (iii) $△\mathrm{ABC}\cong △\mathrm{BAD}$ (iv) diagonal $\mathrm{AC}=$ diagonal BD
   
   Sol. Given :
   
   ABCD is a trapezium. $\mathrm{AB}\Vert \mathrm{CD}$ and $\mathrm{AD}=\mathrm{BC}$ To Prove: (i) $\angle \mathrm{A}=\angle \mathrm{B}$ (ii) $\angle \mathrm{C}=\angle \mathrm{D}$ (iii) $\Delta \mathrm{ABC}\cong △\mathrm{BAD}$ (iv) Diagonal AC = Diagonal BD Construction : Draw CE || AD and extend AB to intersect CE at E. Proof: (i) As AECD is a parallelogram. [By construction] $\therefore \mathrm{AD}=\mathrm{EC}$ But AD = BC [Given] $\therefore \mathrm{BC}=\mathrm{EC}$ $\Rightarrow \angle 3=\angle 4$ [Angles opposite to equal sides are equal] Now, $\angle 1+\angle 4=180^{\circ }$ [Co-Interior angles] And $\angle 2+\angle 3=180^{\circ }$ [Linear pair] $\Rightarrow \angle 1+\angle 4=\angle 2+\angle 3$ $\Rightarrow \angle 1=\angle 2[\because \angle 3=\angle 4]$ $\Rightarrow \angle \mathrm{A}=\angle \mathrm{B}$ (ii) $\angle 3=\angle \mathrm{BCD}$ [Alternate interior angles] $\angle \mathrm{ADC}=\angle 4$ [Opposite angles of a parallelogram] But $\angle 3=\angle 4$ [ $△\mathrm{BCE}$ is an isosceles triangle] $\therefore \angle \mathrm{BCD}=\angle \mathrm{ADC}$ $\therefore \angle \mathrm{C}=\angle \mathrm{D}$ (iii) In $△ABC$ and $△BAD$, $$\begin{gathered} \mathrm{AB}=\mathrm{AB}\text{ [Common] } \\ \angle 1=\angle 2\text{ [Proved] } \\ \mathrm{AD}=\mathrm{BC}[\text{ Given }] \\ \therefore △\mathrm{ABC}\cong △\mathrm{BAD}[\text{ By SAS congruency] } \\ \Rightarrow \mathrm{AC}=\mathrm{BD} \end{gathered}$$
9. The angles of quadrilateral are in the ratio $3:5:9:13$. Find all the angles of the quadrilateral. Sol. Let the four angles of the quadrilateral be $3x,5x,9x$ and $13x$. $3\mathrm{x}+5\mathrm{x}+9\mathrm{x}+13\mathrm{x}=360^{\circ }$ [Sum of all the angles of quadrilateral is $360^{\circ }$ ] $\Rightarrow 30\mathrm{x}=360^{\circ }$ $\Rightarrow \mathrm{x}=12^{\circ }$ Hence, the angles of the quadrilateral are $3\times 12^{\circ }=36^{\circ },5\times 12^{\circ }=60^{\circ }$, $9\times 12^{\circ }=108^{\circ }$ and $13\times 12^{\circ }=156^{\circ }$.
10. Show that if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus. Sol. Given : ABCD is a quadrilateral where diagonals AC and BD meet at 0 , such that $\mathrm{AO}=\mathrm{OC},\mathrm{OB}=\mathrm{OD}$ and $\mathrm{AC}\perp \mathrm{BD}$ To Prove: Quadrilateral ABCD is a rhombus, i.e., $\mathrm{AB}=\mathrm{BC}=\mathrm{CD}=\mathrm{DA}$ Proof: In $△AOB$ and $△AOD$, $OB=OD$ [Given] $\mathrm{AO}=\mathrm{AO}$ [Common] $\angle \mathrm{AOB}=\angle \mathrm{AOD}[$ Each $=90^{\circ }]$ $\therefore △\mathrm{AOB}\cong △\mathrm{AOD}$ [SAS Rule] $\therefore \mathrm{AB}=\mathrm{AD}$ [C.P.C.T.]
    
    Similarly, we can prove that $AB=BC$ $BC=CD$ $CD=AD$ From (i), (ii), (iii) and (iv), we obtain, $\mathrm{AB}=\mathrm{BC}=\mathrm{CD}=\mathrm{DA}$ $\therefore$ Quadrilateral ABCD is a rhombus.
11. Show that if the diagonals of $a$ quadrilateral are equal and bisect each other at right angles, then it is a square. Sol. Given : The diagonals AC and BD of a quadrilateral ABCD are equal and bisect each other at right angles. To prove : Quadrilateral $ABCD$ is a square. Proof:
    
    In $△AOD$ and $△BOC$, $$\begin{gathered} \mathrm{OA}=\mathrm{OC}\text{ [Given] } \\ \mathrm{OD}=\mathrm{OB}\text{ [Given] } \end{gathered}$$ $\angle \mathrm{AOD}=\angle \mathrm{COB}$ [Vertically Opposite Angles] $\therefore △\mathrm{AOD}\cong △\mathrm{COB}$ [SAS Rule] $\therefore \mathrm{AD}=\mathrm{BC}$ [C.P.C.T.] $\angle ODA=\angle OBC$ [C.P.C.T.] $\therefore \mathrm{AD}\Vert \mathrm{BC}$ Now, $\mathrm{AD}=\mathrm{CB}$ and $\mathrm{AD}\Vert \mathrm{CB}$ $\therefore$ Quadrilateral ABCD is a |gm. In $△AOB$ and $△AOD$, $$\begin{gathered} AO=AO[\text{ Common }] \\ OB=OD[\text{ Given }] \\ \angle AOB=\angle AOD\left[\text{ Each }=90^{\circ }(\text{ Given })\right] \\ \therefore \Delta AOB\cong △AOD[\text{ SAS Rule }] \\ \therefore AB=AD \end{gathered}$$ Now, $\therefore \mathrm{ABCD}$ is a parallelogram and $\mathrm{AB}=\mathrm{AD}$ Again, in $△\mathrm{ABC}$ and $△\mathrm{BAD}$, $\mathrm{AC}=\mathrm{BD}$ [Given] $BC=AD[\because ABCD$ is a $\Vert gm]$ $\mathrm{AB}=\mathrm{BA}[$ Common $]$ $\therefore △\mathrm{ABC}\cong △\mathrm{BAD}[\mathrm{SSS}$ rule] $\therefore \angle \mathrm{ABC}=\angle \mathrm{BAD}$ [C.P.C.T.] $\because \mathrm{AD}|\mid \mathrm{BC}$ [Opposite sides of ||gm ABCD] and transversal AB intersects them. $\therefore \angle \mathrm{ABC}+\angle \mathrm{BAD}=180^{\circ }$ [Sum of consecutive interior angles on the same side of the transversal is $180^{\circ }$ ] $\therefore \angle \mathrm{ABC}=\angle \mathrm{BAD}=90^{\circ }$ Similarly, $\angle \mathrm{BCD}=\angle \mathrm{ADC}=90^{\circ }$ $\therefore \mathrm{ABCD}$ is a square.
12. $ABCD$ is a rhombus. Show that diagonal AC bisects $\angle \mathrm{A}$ as well as $\angle \mathrm{C}$ and diagonal BD bisects $\angle \mathrm{B}$ as well as $\angle \mathrm{D}$. Sol Given : ABCD is a rhombus and AC and BD are its diagonals. To prove : (i) Diagonal AC bisects $\angle \mathrm{A}$ as well as $\angle \mathrm{C}$. (ii) Diagonal BD bisects $\angle \mathrm{B}$ as well as $\angle \mathrm{D}$.
    
    Proof: (i) In $△\mathrm{ABC}$ $\mathrm{AB}=\mathrm{BC}$ (sides of Rhombus) So, $\angle 2=\angle 4$ (Angle opposite to equal sides are equal) But $\angle 2=\angle 3$ (Alternate angles as $\mathrm{AB}\Vert \mathrm{CD}$ ) So, $\angle 2=\angle 3=\angle 4$ But $\angle 1=\angle 4$ (Alternate angles as $\mathrm{AD}\Vert \mathrm{BC}$ ) So, $\angle 1=\angle 2=\angle 3=\angle 4$ $\angle 1=\angle 2$ by (1) So, AC bisect $\angle \mathrm{A}$ $\angle 3=\angle 4$ by (1) So, AC bisect $\angle \mathrm{C}$ (ii) In $△ABD$
    
    $\mathrm{AB}=\mathrm{AD}$ (Sides of Rhombus) So, $\angle 5=\angle 7$ (Angle opposite to equal sides are equal) But $\angle 7=\angle 6$ (Alternate angles as $\mathrm{AD}\Vert \mathrm{BC}$ ) So, $\angle 5=\angle 6=\angle 7$ $\angle 5=\angle 8$ (Alternate angles as $\mathrm{AB}\Vert \mathrm{CD}$ ) So, $\angle 5=\angle 6=\angle 7=\angle 8$ $\angle 5=\angle 6$ by (2) So, BD bisect $\angle \mathrm{B}$ $\angle 7=\angle 8$ by (2) So, BD bisect $\angle \mathrm{D}$
13. In $△\mathrm{ABC}$ and $△\mathrm{DEF},\mathrm{AB}=\mathrm{DE},\mathrm{AB}\Vert \mathrm{DE},\mathrm{BC}=$ EF and BC | EF. Vertices A, B and C are joined to vertices $\mathrm{D},\mathrm{E}$ and F respectively. Show that : (i) quadrilateral ABED is a parallelogram (ii) quadrilateral BEFC is a parallelogram (iii) $\mathrm{AD}\Vert \mathrm{CF}$ and $\mathrm{AD}=\mathrm{CF}$ (iv) quadrilateral ACFD is a parallelogram (v) $\mathrm{AC}=\mathrm{DF}$ (vi) $△\mathrm{ABC}\cong △\mathrm{DEF}$
    
    Sol. Given : $\mathrm{AB}=\mathrm{DE},\mathrm{AB}\Vert \mathrm{DE},\mathrm{BC}=\mathrm{EF}$ & $\mathrm{BC}\Vert \mathrm{EF}$ To Prove (i) ABED is a parallelogram. (ii) BEFC is a parallelogram. (iii) $\mathrm{AD}\Vert \mathrm{CF}$ and $\mathrm{AD}=\mathrm{CF}$ (iv) ACFD is a parallelogram (v) $\mathrm{AC}=\mathrm{DF}$ (vi) $△\mathrm{ABC}\cong △\mathrm{DEF}$ Proof (i) In quad. ABED $\mathrm{AB}=\mathrm{DE}$ And AB || DE [Given] $\therefore$ ABED is a parallelogram (ii) In quad. BEFC $\mathrm{BC}=\mathrm{EF}$ [Given] And $\mathrm{BC}\mid \mathrm{EF}$ [Given] $\therefore BEFC$ is a parallelogram. (iii) As ABED is a parallelogram. $\therefore \mathrm{AD}\Vert \mathrm{BE}$ and $\mathrm{AD}=\mathrm{BE}$ Also, BEFC is a parallelogram $\therefore \mathrm{CF}\Vert \mathrm{BE}$ and $\mathrm{CF}=\mathrm{BE}$ From (i) and (ii), we get $\therefore \mathrm{AD}\Vert \mathrm{CF}$ and $\mathrm{AD}=\mathrm{CF}$ (iv) $\mathrm{As}\mathrm{AD}\Vert \mathrm{CF}$ and $\mathrm{AD}=\mathrm{CF}$ $\Rightarrow$ ACFD is a parallelogram. (v) As ACFD is a parallelogram. $\therefore \mathrm{AC}=\mathrm{DF}$ (vi) In $△\mathrm{ABC}$ and $△\mathrm{DEF}$, AB $=\mathrm{DE}$ $\mathrm{BC}=\mathrm{EF}$ [Given] $\mathrm{AC}=\mathrm{DF}$ $[$ Given] [Proved] $\therefore △\mathrm{ABC}\cong △\mathrm{DEF}$ [By SSS congruency]

4.0Exercise: 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 (fig.) and AC is a diagonal. Show that (i) $\mathrm{SR}\Vert \mathrm{AC}$ and SR $=\frac{1}{2}\mathrm{AC}$ (ii) $PQ=SR$ (iii) $PQRS$ is a parallelogram.
   
   Sol. Given : ABCD is a quadrilateral in which $P,Q,R$ and $S$ are mid-points of $AB,BC,CD$ and $DA.AC$ is a diagonal. To prove : (i) $\mathrm{SR}\Vert \mathrm{AC}$ and $\mathrm{SR}=\frac{1}{2}\mathrm{AC}$ (ii) $PQ=SR$ (iii) $PQRS$ is a parallelogram. Proof: (i) $\mathrm{In}△\mathrm{DAC}$, $\because S$ is the mid-point of $DA$ and $R$ is the mid-point of DC $\therefore \mathrm{SR}\Vert \mathrm{AC}$ and $\mathrm{SR}=\frac{1}{2}\mathrm{AC}$ [By Mid-point theorem] (ii) In $△BAC$, $\because P$ is the mid-point of $AB$ and $Q$ is the mid-point of $BC$ $\therefore \mathrm{PQ}\Vert \mathrm{AC}$ and $\mathrm{PQ}=\frac{1}{2}\mathrm{AC}$ [By Mid-point theorem] But from (i) $\mathrm{SR}=\frac{1}{2}\mathrm{AC}\&(\mathrm{ii})$ $PQ=\frac{1}{2}AC$ $\Rightarrow \mathrm{PQ}=\mathrm{SR}$ (iii) $\mathrm{PQ}\Vert \mathrm{AC}[$ From (ii)] SR | AC [From (i)] $\therefore \mathrm{PQ}\Vert \mathrm{SR}$ [Two lines parallel to the same line are parallel to each other] Also, $\mathrm{PQ}=\mathrm{SR}$ [From (ii)] $\therefore$ PQRS is a parallelogram. [A quadrilateral is a parallelogram if a pair of opposite sides is parallel and is of equal length]
2. $ABCD$ is a rhombus and $P,Q,R$ and $S$ are the mid points of sides $\mathrm{AB},\mathrm{BC},\mathrm{CD}$ and DA respectively. Show that the quadrilateral PQRS is a rectangle. Sol.
   
   Given : $P,Q,R$ and $S$ are the mid-points of respective sides $AB,BC,CD$ and $DA$ of rhombus. $\mathrm{PQ},\mathrm{QR},\mathrm{RS}$ and SP are joined. To prove : PQRS is a rectangle. Construction : Join A and C. Proof: In $△ABC,P$ is the mid-point of $AB$ and $Q$ is the mid-point of $BC$. $\therefore \mathrm{PQ}\Vert \mathrm{AC}$ and $\mathrm{PQ}=\frac{1}{2}\mathrm{AC}$ (By midpoint theorem) In $△ADC,R$ is the mid-point of $CD$ and $S$ is the mid-point of $AD$. $\therefore \mathrm{SR}\Vert \mathrm{AC}$ and $\mathrm{SR}=\frac{1}{2}\mathrm{AC}$. (By midpoint theorem) From eq. (i) and (ii), $PQ\Vert SR$ and $PQ=SR$ $\therefore \mathrm{PQRS}$ is a parallelogram. Now ABCD is a rhombus [Given] $\therefore \mathrm{AB}=\mathrm{BC}$ $\Rightarrow \frac{1}{2}\mathrm{AB}=\frac{1}{2}\mathrm{BC}\Rightarrow \mathrm{PB}=\mathrm{BQ}$ $\therefore \angle 1=\angle 2$ [Angles opposite to equal sides are equal] Now in triangles APS and CQR, we have, $\mathrm{AP}=\mathrm{CQ}$ [ P and Q are the mid-points of AB and BC and $AB=BC]$ Similarly, AS = CR and PS = QR [Opposite sides of a parallelogram] $\therefore \Delta \mathrm{APS}\cong △\mathrm{CQR}$ [By SSS congruency] $\Rightarrow \angle 3=\angle 4$ [By C.P.C.T.] Now, we have $\angle 1+\angle \mathrm{SPQ}+\angle 3=180^{\circ }$ And $\angle 2+\angle \mathrm{PQR}+\angle 4=180^{\circ }$ $\therefore \angle 1+\angle \mathrm{SPQ}+\angle 3=\angle 2+\angle \mathrm{PQR}+\angle 4$ Since $\angle 1=\angle 2$ and $\angle 3=\angle 4$ [Proved above] $\therefore \angle \mathrm{SPQ}=\angle \mathrm{PQR}$ Now PQRS is a parallelogram [Proved above] $\therefore \angle \mathrm{SPQ}+\angle \mathrm{PQR}=180^{\circ }$ [Co-Interior angles] Using eq. (iii) and (iv), $\angle \mathrm{SPQ}+\angle \mathrm{SPQ}=180^{\circ }$ $\Rightarrow 2\angle \mathrm{SPQ}=180^{\circ }$ $\Rightarrow \angle \mathrm{SPQ}=90^{\circ }$ Hence, $PQRS$ is a rectangle.
3. $ABCD$ is a rectangle and $P,Q,R$ and $S$ are mid-points of the sides $\mathrm{AB},\mathrm{BC},\mathrm{CD}$ and DA respectively. Show that the quadrilateral $PQRS$ is a rhombus. Sol. Given : A rectangle ABCD in which $\mathrm{P},\mathrm{Q},\mathrm{R}$ and $S$ are the mid-points of the sides $AB$, $\mathrm{BC},\mathrm{CD}$ and DA respectively. PQ, QR, RS and SP are joined. To prove : PQRS is a rhombus.
   
   Construction : Join AC. Proof : In $△\mathrm{ABC},\mathrm{P}$ and Q are the midpoints of sides $\mathrm{AB},\mathrm{BC}$ respectively. $\therefore \mathrm{PQ}\Vert \mathrm{AC}$ and $\mathrm{PQ}=\frac{1}{2}\mathrm{AC}$ [Mid- Point Theorem] In $△\mathrm{ADC},\mathrm{R}$ and S are the mid-points of sides $\mathrm{CD},\mathrm{AD}$ respectively. $\therefore \mathrm{SR}\Vert \mathrm{AC}$ and $\mathrm{SR}=\frac{1}{2}\mathrm{AC}$ [Mid-Point Theorem] From eq.(i) and (ii), $PQ\Vert SR$ and $\mathrm{PQ}=\mathrm{SR}$ $\therefore PQRS$ is a parallelogram. Now ABCD is a rectangle. $\therefore \mathrm{AD}=\mathrm{BC}$ $\Rightarrow \frac{1}{2}\mathrm{AD}=\frac{1}{2}\mathrm{BC}$ $\Rightarrow \mathrm{AS}=\mathrm{BQ}$ In triangles APS and BPQ, $\mathrm{AP}=\mathrm{BP}[\mathrm{P}$ is the mid-point of AB ] $\angle \mathrm{PAS}=\angle \mathrm{PBQ}[$ Each $90^{\circ }]$ And AS = BQ [From eq. (iv)] $\therefore △\mathrm{APS}\cong △\mathrm{BPQ}$ [By SAS congruency] $\Rightarrow \mathrm{PS}=\mathrm{PQ}$ [By C.P.C.T.] From eq.(iii) and (v), we get that PQRS is a parallelogram. $\Rightarrow PS=PQ$ $\Rightarrow$ Two adjacent sides are equal. Hence, PQRS is a rhombus.
4. ABCD is a trapezium in which $\mathrm{AB}\Vert \mathrm{DC},\mathrm{BD}$ is a diagonal and E is the mid-point of AD . A line is drawn through E parallel to $AB$ intersecting BC at F (fig.). Show that F is the mid-point of BC .
   
   Sol. Line $\ell \Vert \mathrm{AB}$ and passes through E.
   
   Line $\ell$ meets BC at F and BD at G . In $△ABD,E$ is mid-point of $AD$ and $EG\Vert AB$. $\Rightarrow G$ is mid-point of $BD$. [Converse of Mid Point Theorem] Also, $\ell \Vert \mathrm{AB}$ and $\mathrm{AB}\Vert \mathrm{CD}$ $\Rightarrow \ell \Vert CD$ $\Rightarrow$ Fis mid-point of $BC$. $[\because G$ is mid-point of $BD]$ [Converse of Mid Point Theorem]
5. In a parallelogram $\mathrm{ABCD},\mathrm{E}$ and F are the mid-points of sides AB and CD respectively (fig.). Show that the line segments AF and EC trisect the diagonal BD .
   
   Sol. Given : ABCD is a parallelogram. E and F are midpoints of $AB$ and $DC$ respectively.
   
   To prove: $\mathrm{DP}=\mathrm{PQ}=\mathrm{QB}$ Proof: Since E and F are the mid-points of $AB$ and $CD$ respectively. $\therefore \mathrm{AE}=\frac{1}{2}\mathrm{AB}$ and $\mathrm{CF}=\frac{1}{2}\mathrm{CD}$ But ABCD is a parallelogram. $\therefore \mathrm{AB}=\mathrm{CD}$ and $\mathrm{AB}\Vert \mathrm{DC}$ $\Rightarrow \frac{1}{2}\mathrm{AB}=\frac{1}{2}\mathrm{CD}$ and $\mathrm{AB}\Vert \mathrm{DC}$ $\Rightarrow \mathrm{AE}=\mathrm{FC}$ and $\mathrm{AE}\Vert \mathrm{FC}$ [From eq. (i)] $\therefore$ AECF is a parallelogram. $\Rightarrow \mathrm{FA}\Vert \mathrm{CE}$ $\Rightarrow \mathrm{FP}\Vert \mathrm{CQ}$ [FP is a part of $FA$ and $CQ$ is a part of CE] ... (ii) Since the line segment drawn through the mid-point of one side of a triangle and parallel to the other side bisects the third side. In $△DCQ,F$ is the mid-point of $CD$ and $\Rightarrow \mathrm{FP}\Vert \mathrm{CQ}$ $\therefore \mathrm{P}$ is the mid-point of DQ . $\Rightarrow DP=PQ$ Similarly, In $△ABP,E$ is the mid-point of AB and $\Rightarrow \mathrm{EQ}\Vert \mathrm{AP}$ $\therefore \mathrm{Q}$ is the mid-point of BP . $\Rightarrow BQ=PQ$ From eq.(iii) and (iv), $DP=PQ=BQ$ Now, $BD=BQ+PQ+DP=BQ+BQ+BQ=3BQ$ $\Rightarrow \mathrm{BQ}=\frac{1}{3}\mathrm{BD}$ From eq (v) and (vi), $DP=PQ=BQ=\frac{1}{3}BD$ $\Rightarrow$ Points P and Q trisects BD . So, AF and CE trisects BD. Hence, AF and CE trisect the diagonal BO.
6. 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) $\mathrm{MD}\perp \mathrm{AC}$ (iii) $\mathrm{CM}=\mathrm{MA}=\frac{1}{2}\mathrm{AB}$ Sol. (i) Through M, we draw line $\ell \Vert BC$. $\ell$ intersects AC at D. $M$ is a midpoint of $AB$ $\Rightarrow D$ is mid-point of $AC$. [By converse of mid-point theorem]
   
   (ii) $\angle \mathrm{ADM}=\angle \mathrm{ACB}=90^{\circ }$ [Corresponding angles] $\Rightarrow \angle \mathrm{ADM}=90^{\circ }$ $\Rightarrow \mathrm{MD}\perp \mathrm{AC}$. (iii) In $△$ CMD and $△\mathrm{AMD}$; $CD=AD,MD=MD$ and $\angle \mathrm{CDM}=\angle \mathrm{ADM}$ $[$ Each $=90^{\circ }$ ] Therefore, $△\mathrm{CMD}\cong △\mathrm{AMD}$ [SAS congruence rule] $\Rightarrow \mathrm{CM}=\mathrm{AM}$; Also $\mathrm{AM}=\frac{1}{2}\mathrm{AB}$ So, $CM=\frac{1}{2}AB$ $\mathrm{AM}=\mathrm{CM}=\frac{1}{2}\mathrm{AB}$
7. Show that the line segments joining the mid-points of the opposite sides of a quadrilateral bisect each other. Sol. P,Q,R and $S$ are the mid-points of the sides $AB,BC,CD$ and $AD$ of the quadrilateral ABCD.
   
   We have to prove that, $PR$ and $QS$ bisects each other. Now, join PQ, QR, RS and PS. Here, we can prove that $PQRS$ is a parallelogram (as in solution 1). Now, PR and QS are the diagonals of the parallelogram PQRS. $\therefore \mathrm{PR}$ and QS bisect each other as diagonals of parallelogram bisect each other.