Exercise 11 B
Question 1
(IMAGE TO BE ADDED)
(i) ABCD is a parallelogram in which diagonal AC = Diagonal
BD and AC and BD intersect each other at O at right angle
To prove: ABCD is square
Proof: In $\triangle A B C$ and $\triangle A D B$
AB= AB
AC = BD
BC = AD
So $\triangle A B C \cong \triangle A D C$
So $\angle A B C=\angle B A D$
But $\angle A B C+\angle B A D=180^{\circ}$
So $\angle A B C=\angle B A D=90^{\circ}$
Again in $\triangle A O B$ and $\triangle B O C$
$\begin{aligned}&\angle A O B=\angle B O C \\&B O=B O \\&A O=O C\end{aligned}$
So $\triangle A O B \cong \triangle B O C$
So $A B=B C$
But $A B=C D$ and $B C=A D$
So $A B=B C=C D=D A$
And each angle is $90^{\circ} .$
Hence ABCD is a square
Hence proved
(ii) (IMAGE TO BE ADDED)
ABCD is a rectangle and AC and BD arc its diagonals
To prove: AC=BD
Proof: In$\triangle A B C$ and $\triangle A D B, A B=A B$
$B C=A D$
$\angle A B C=\angle B A D$
So $\triangle A B C \cong \triangle A D B$
So $A C=B D$ Hence proved
Question 2
Sol: In parallelogram ABCD , AC is its diagonal X and Y are the mid points of AB and CD respectively
M is mid point of AC and AX = CY
To prove:
(i) $\triangle A X M \cong \triangle C Y M$
(ii) XMY is a straight line
Construction : Join XM and YM
Proof : In $\triangle$ A M X and $\triangle C YM $
$A X=C Y$
$A M=C M $
And $\angle 4 \mathrm{~cm}=\angle M A X$
So
$\begin{aligned} \triangle A M X & \cong \triangle CMY \\ \text { So } A M X &=\angle CMY \end{aligned}$
But these are vertically opposite angles
So XMY is a straight line
Hence proved '
Question 3
Sol:(i) In a square ABCD , BD is its diagonal
To Prove: $\angle A B D=45^{\circ}$
Proof : In $\triangle A B D$ and $\triangle B C D$
AB = DC
AD= BC
BD =BD
So $\triangle A B D \cong \triangle B C D$
So $\angle A B D=\angle C B D$
But $\angle A B D+\angle C B D=90^{\circ}$
$\angle A B D=\frac{90^{\circ}}{2}=45^{\circ}$
Hence diagonal BD makes an angle of $45^{\circ}$ with the side of the square
(ii) (IMAGE TO BE ADDED)
In rhombus ABCD its diagonals AC and BD bisects each other at O
To Prove: $\angle A O B=\angle B O C=\angle C O D$
$\angle D O A=90^{\circ}$
Proof: In $\triangle A O B$ and $\triangle C O B$
$A O=O C$
$B O=B O$
$A B=B C$
So $\triangle A O B \cong \triangle C O B$
So $\angle A O B=\angle C O B$
But $\angle A O B+\angle C O B=180^{\circ}$
$\angle A O B=\angle C O B=90^{\circ}$
Similarly we can prove that
$\angle C O D=\angle D O A=90^{\circ}$
Hence diagonals of a rhombus bisect each other at Right angles
(iii) In rhombus ABCD, AC is its Diagonals
To prove: AC bisect $\angle A$ and $\angle C$
Proof: In $\angle A B C$ and $\triangle A D C$
$A B=C D$
$B C=A D$
$A C=A C$
So $\triangle A B C \cong \triangle A D C$
So $\angle C A B=\angle C A D$
and $\angle A C B=\angle A C D$
Hence diagonal of a rhombus bisects the angles at the vertices
Question 4
PQRS is a parallelogram in which PR is its diagonal ST
$\perp P R$ and $Q U \perp P R$
To prove :
(i) $\Delta S T R \cong \Delta Q U P \quad$ (ii) $S T=Q U$
Proof : In $\triangle$ STR and $\triangle Q U P$
$P S=S R$
$\angle T=\angle U$
$\angle S P T=\angle U R Q$
(i) So $\triangle$ STR $\cong \triangle Q U P$
(ii) so $S T=Q U$
Question 5
Sol: In ||gm ABCD , PO and QO are the angle bisectors of $\triangle P$ and $\triangle Q$ respectively meeting Each other at O . $\triangle QM$ is drawn P parallel to PQ through O
To prove:
(i) $P L=Q P \quad$ (iii L O=O M$
Proof : If PO is the bisects of $\angle P$
So $\angle 1=\angle 2$
Similarly QO is the bisector of $\angle Q$
so $\angle 5=\angle 6$
if $\angle M \| P Q$
$\angle PQM is a parallelogram
And $\angle 3=\angle 2=\angle 1$
And $\angle 4=\angle 5=\angle 6$
(i) LP = QM
(ii) PL = LO
and OM = OM
But PL = OM
OL =OM
Hence proved
Question 6
(i) Sol: In ||gm ABCD, BD is its diagonal L and M are points on AB and CD respectively such that AL =CM
To prove: LM and BD bisect each other
i.e BO =OD and LO = OM
Proof : In || gm ABCD,
AB=CD
But AL = CM
So AB-AL = CD - CM
=LM =MD
Now in $\triangle L O B$ and $\triangle D O M$
$\angle B=M D$
$\angle L B O=\angle M D O$
$\angle L O B=\angle D O M$
So $\triangle L O B \cong \triangle D O M$
SO $O B=O D$ and $O L=O M $
Hence BD and LM bisect each other
(ii) In a parallelogram ABCD, AB is produced to E such that BE = AB
To prove: ED bisects BC
Proof: IF AB= DC
and AB= BE
DC = BE
Now in $\triangle B O E$ and $\triangle C O D$
BE=DC
$\begin{aligned} \angle B O E &=\angle C O D \\ \angle \angle B E O &=\angle C D O \\ \text { SO } \triangle B O E & \cong \triangle C O D \\ \text { SO } B O &=O C \end{aligned}$
Hence DE bisects BC
Question 7
(IMAGE TO BE ADDED)
(i) In parallelogram ABCD diagonals AC and BD intersect each other at O.
A line XOY is drawn which meets AB in X and CD in Y
To prove: OX =OY
Proof: In $\triangle X O B$ and $\triangle Y O D$
$O B=O D$
$\angle B O X=\angle D O Y$
$\angle O B X=\angle O D Y$
So $\triangle X O B \cong \triangle Y O D$
So $O X=O Y$
Hence proved
(ii) In $\triangle A B C, A D$ is the median which is produced to X such that DX =AB BX and XC are joined
To Prove: ABXC is a parallelogram
Proof: In $\triangle A D B$ and $\triangle X D C$
$A D=D X$
$B D=D C$
$\angle A D B=\angle X D C$
So $\triangle A D B \cong \triangle \X D C$
SO $A B=C X$
$ \angle B A D=\angle C X D$
But these are alternate angles
So AB||XC
If AB =XC and AB||XC
So ABXC is a parallelogram (HENCE PROVED)
Question 8
Sol: (IMAGE TO BE ADDED)
(i) In parallelogram ABCD, AE is the bisector of $\angle A$ which meets CD in E. AB = 2AD
ED is joined
To prove:
(i) BE bisects LE
(ii)$\angle A E B$ is a right angle
Construction: Take F , a mid point of AB
So AF =AD
If $\angle 1=\angle 2$
So AFED is a rhombus
$A D\|E F\| C D$ and E is mid point of CD
So FBCE is also a rhombus
Whose BE is diagonal
But diagonal of a rhombus Bisects the angles at the vertices
So BE bisects LE
(ii) SO $\angle 1=\angle 2$ and $\angle 3=\angle 4$
But $\angle A+\angle B=180^{\circ}$
So $2 \angle 2+2 \angle 3=180^{\circ}$
$\Rightarrow \angle 2+\angle 3=90^{\circ}$
so In $\triangle A E B$,
if $\begin{aligned} & \angle 2+\angle 3+\angle A E B=180^{\circ} \\ \Rightarrow & 90^{\circ}+\angle A E B=180^{\circ} \\ \Rightarrow & \angle A E B=180^{\circ}-90^{\circ} \end{aligned}$
$\Rightarrow \angle A E B=90^{\circ}$
Hence $\angle A E B$ is a right angle
Question 9
(IMAGE TO BE ADDED)
Sol: (i) In ||gm ABCD,
X and Y are mid points of the side AD and BC
respectively . BD is its diagonals AY and CX are joined
To prove:
(i) ATCX is a ||gm
(ii) XY and BD bisect each other
Proof : In AB||BC
= AX ||CY
But X and Y are mid point of AD and BC respectively and AD = BC
AX=CY
so AXCY is a ||gm
(ii) In $\triangle X O D$ and $\triangle B O Y$
$X D=B Y$
$\angle X O D=\angle B O Y$
$\angle A D O=O B Y$
So $ \triangle X O D \cong \triangle B O Y$
XO =OY
and DO= OB
Hence XY and BD Bisect Each other (Hence proved)
(ii) In ||gm ABCD ,
BM is the bisector of $\angle B$ and DN is the bisector of $\angle D$
To prove:
(i) BNDM is a ||gm
(ii) BM=ON
Proof: If bisector of $\angle A B C$ Meets AD at M
So M is mid point of AD
Similarly DN is the bisector of $\angle A D C$
So N is mid point of BC
So MD||BN
and DM||BN
so BNDM is a ||gm
So BM =DM
Hence proved
(iii) In ||gm ABCD, BM is the bisector of $\angle B$ and AN is the bisector of $\angle A$
To prove:
(i) $M N=C D$
(ii) ABNM is a rhombus
Proof: $\angle A+\angle B=180^{\circ}$
If AN and BM are the bisector of $\angle A$ and $\angle B$ respectively
So $\begin{aligned} \frac{1}{2} \angle A+\frac{1}{2} \angle B &=90^{\circ} \\ \Rightarrow \angle O A B+\angle D B A &=90^{\circ} \end{aligned}$
So In $\triangle A O B$,
$\angle A O B=180^{\circ}-90^{\circ}=90^{\circ}$
So AN and BM are perpendicular to each other Now in $\triangle A O B$ and $\triangle M O N$
$A B=M N$
$\angle A O B=\angle M O N$
$\angle B A N=\angle A N M$
SO $\triangle A O B \cong \triangle M O N$
SO $A O=O N$ and $B O=0 M$
If diagonals BM and AN bisects each other at right angles.
So ABNM is a rhombus.
Question 10
(IMAGE TO BE ADDED)
Sol: Two parallel line PQ and RS and a transversal LM intersect them at A and B respectively
This bisectors of interior angles at A and B, Meet each other at C and D as shown in the figure forming a quadrilateral ABCD
To prove:
(i) ABCD is a rectangles
(ii) CD is parallel to the original parallel lines PQ and RS
Construction: Join CD
Proof:
(i) AC and BD are the bisector of $\angle P A B$ and $\angle A B S$ and AC||BD
Similarly AD||BC
ACBD is a parallelogram
Now $\angle P A B+\angle A B R=180^{\circ}$
So $\frac{1}{2} \angle P A B+\frac{1}{2} \angle A B R=90^{\circ}$
$\Rightarrow \angle C A B+\angle C B A=90^{\circ}$
So In $\triangle A B C$
$\angle A C B=90^{\circ}$
So A parallelogram with each angle a right angle is a rectangle
Hence ABCD is a rectangle
(ii) If diagonals of a rectangle are equal and bisect each other
Diagonal AB and CD of rectangle ABCD Bisects each other At O
So AB = OC
So $\angle O A C=\angle A C O \Rightarrow C C A P=\angle A C O$
But these are alternate angels
So $C D \| P Q$ or $R S$
Question 11
Sol: (i) In $\triangle A B C, B Y$ is the bisector of Meeting AC at Y
Through Y, YX ||AB and YZ ||BC are drawn
TO prove: BXYX is a rhombus
Proof: if XY||BA OR BZ and YZ||BC or BX
(IMAGE TO BE ADDED)
So BZYX is parallelogram
But BY is the diagonal which bisects $\angle B$.
So $B Z Y X$ is a rhombus
(ii)HJKL is a square in which diagonals HK and JL Intersect each other at O
X is a point on HJ such that HX =HO
XO is joined
To prove: $\angle H O X=3 \angle X O J$
Proof : If diagonals of square HJKL Bisects each other at right angle at O
So $\angle H O J=90^{\circ}$ and $\angle K H J$ or $\angle O H X=45^{\circ}$
If In $\Delta O H X$
OH=HX
so $\angle H O X=\angle H XO$
But $\angle O H X+\angle H O X+\angle H X O=180^{\circ}$
$\Rightarrow 45^{\circ}+\angle H O X+\angle H O X=180^{\circ}$
$\Rightarrow 2 \angle H O X=180^{\circ}-45^{\circ}=135^{\circ}$
$\Rightarrow \angle H O X=\frac{135}{2}=67 \frac{1}{2}^{\circ}$.........(i)
$\begin{aligned} \text { But } \angle X O T &=\angle H O J=\angle H O X \\ 90^{\circ}-67 \frac{1^{\circ}}{2} &=22 \frac{1^{\circ}}{2} \end{aligned}$...........(ii)
From (i) and (ii)
$\angle H O X=67 \frac{1}{2}^{\circ}=3 \times 22 \frac{1}{2}$
3 $\angle XOJ $
(iii) In the figure
ABCD is ||gm
ABL and ADXY are square
To prove : $\triangle $ CXM is an isosceles triangle
Construction: Join MC, MX and CX
Proof : In parallelogram ABCD
AB=CD and BC =AD
and $\angle A B C=\angle C D A$ and $\angle B C D=\angle B A D$
So $\angle M B C=\angle M B A+\angle A B C=90^{\circ}+\angle A B C$.............(i)
$\angle C D X=\angle C D A+\angle A D X$..............(ii)
$=\angle A B C+90^{\circ}$
So From (i) and (ii)
$\angle M B C=\angle C D X$
Now in $\triangle B M C$ and $\triangle D C X$
BM =BA =CD
BC =AD = DX
and $\angle M B X=\angle C D X$
So $\triangle B A C \cong \triangle D C X$
So $CM=CX$
Hence $\triangle CMX$ is an isosceles triangle
(iv) ABCD and ALMN are two square as shown in the figure
(IMAGE TO BE ADDED)
To prove: BL=DN
Construction : Join DM and BL
Proof: In $\triangle A D N$ and $\triangle A L B$
$A D=A B$
$A N=A L$
$\angle D A N=\angle L A B$
SO $\triangle A D N \approx \triangle A L B$
SO $D N=B L$
Hence proved
Question 12
Sol: (i) In ||gm ABCD, AE and BE are the bisector of adjacent $\angle A$ and $\angle B$ which mect at $E$
To prove: $\angle E$ is a right angle
Proof : If AE is the bisector of $\angle A$
So $\angle E A B=\frac{1}{2} \angle A$
Similarly BE is the bisector of $\angle B$
$\angle E B A=\frac{1}{2} \angle B$
But $\angle A+\angle B=180^{\circ}$
So $\frac{1}{2} \angle A+\frac{1}{2} \angle B=90^{\circ}$
$\angle E A B+\angle E B A=90^{\circ}$
But in $\triangle A E B$
$\angle E A B+\angle E B A+\angle A E B=180^{\circ}$
$\Rightarrow 90^{\circ}+\angle A E B=180^{\circ}$
$\Rightarrow \angle A E B=180^{\circ}-90^{\circ}=90^{\circ}$
So $\angle E$ is a right angle
( ii) In ||gm ABCD PA, QD, RC and SB are the bisector of
$\angle P, \angle Q, \angle R$ and $\angle S$ Respectively Which meet each others
At A,C,D and B respectively forming a quad, ABCD
To prove: ABCD is rectangle
Proof: If PD and QD are the bisector of two adjacent angles $\angle P$ and $\angle Q$ Respectively
So $\quad \angle D=90^{\circ}$
Similarly $\angle B=90^{\circ}$
And PA and SA are the bisector of $\angle P$ and $\angle S$ respectively
So $\angle A=90^{\circ}$
Similarly $\angle C=90^{\circ}$
So the quad. ABCD has its each angle equal to 90
So ABCD is a rectangle Hence proved
Question 13
Sol: ABCD is a ||gm and X is mid point of BC AX is joined and produced to meet BC produced at Q. The ||gm ABPQ is completed as shown in the figure
To prove:
(i)$\triangle A B X \cong \triangle Q C X$
(ii)DC =CQ = QP
Proof: IN $\triangle A B X$ and $\triangle Q C X$
$B X=X C$
$\angle A \times B=\angle C \times Q$
$\angle A B X=\angle X C Q$
So $\triangle A D X \cong \triangle Q C X$
So AB=CQ
But AB = DC and AB =QP
So DC =CQ = QP
Hence proved
Question 14
Sol: P,Q and R are the mid points of the sides AB,BC and CD of rhombus ABCD PQ and QR are joined
To prove: $P Q \perp Q R$
Construction: Join diagonals AC and BD of the rhombus ABCD
Proof: In $\triangle A B C$
P and Q are mid points of AB and BC respectively
So
PQ||AC ...................(i)
Similarly in $\triangle B C D$
Q and R the mid points of BC and CD respectively
So $Q R \quad \| B D$ ................(ii)
But AC and BD bisects each other at right angles
$P Q \perp Q R$
(IMAGE TO BE ADDED)
Question 15
Sol: In a rhombus ABCD
RABS is straight line such that RA = AB = RS
RD and SC are joined and produced to meets at T
To prove: $\angle T=90^{\circ}$
Proof: In $\triangle R A D$
RA =AB = AD
so $\angle D R A=\angle R D A$
and $Ext. \cdot \angle D A B=\angle D R A+\angle R D A$
$=\angle D R A+\angle D R A=2 \angle D R A$..........(i)
Similarly in $\triangle B S C$
BS = AB = BC
So $\angle B C S=\angle B S C$
and Ext. $\angle A B C=\angle B C S+\angle B S C$
$=\angle B S C+\angle B S C=2 \angle B S C$.........(ii)
But in rhombus ABCD
$\angle D O B+\angle A B C=2 \angle B S C$............(iii)
$\angle D O B+\angle A B C=180^{\circ}$
2 $\angle D R A+2 \angle B S C=180^{\circ}$
$\angle D R A+\angle B S C=90^{\circ}$
Now in $\triangle R T S$.
$\angle D R A+\angle B S C=90^{\circ}$
Or $\angle T R S+\angle R S T=90^{\circ}$
$\angle R T S=180^{\circ}-90^{\circ}=90^{\circ}$
Hence $\angle T=90^{\circ}$
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