Exercise 8 A
Question 1
Sol: Two lines segments AB and CD bisect each other at K
AC and BD are joined
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PROVE:AC=BD
In $\triangle A K C$ and $\triangle B K D$
$A K=B K$
$C K=D K$
$\angle A K C=\angle B K D$
So $\triangle A K C \cong \triangle B K D$
So $A C=B D$ Hence proved
Question 2
Sol: In $\triangle A B C$, sides $B A$ and $C A$ are produced such that $B A=A D$ and $C A=A E$. ED is Joined
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To prove:$D E \| B C$
In $\triangle B A C$ And $\triangle D A E$
$B A=D A$
$C A=E A$
$\angle B A C=\angle D A E$
So
$\triangle B A C \cong \triangle D A E$
So $\angle A B C=\angle A D E$
But these are alternate angles
So
$D E \| B C$ Proved
Question 3
Sol: Figure , ABCD is a rectangle p is the mid point of AB.Q and R are points in AD and BC Respectively such that
PQ and PR are joined
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prove $\because P \Q =P R$
$\triangle P A Q$ and $\triangle P B R$
$P A=P B$(∵P is mid point of AB)
$\angle A=\angle B \quad$ (eatch $90^{\circ}$ )
$A Q=B R$
SO $\triangle P A Q \cong \triangle P B R$
$P Q=P R$ poved
Question 4
Sol: $O A=O B$
$O C=O D: \angle A O B=\angle C O D$
To proved
$A C=B D$
$\angle A O B=\angle C O D$
So $\angle A O C+\angle C O B=\angle D O B+\angle C D B$
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$\angle A O C=\angle B O D$
$\triangle O A C$ and $\triangle O D B$
$O A=O B$
$O C=O D$
and $\angle A O C=\angle B O D$
So $\triangle O A C \cong \triangle O D B$
So $A C=B D$
proved
Question 5
Sol:In the figure $O A=O B$
AM $\perp XY$ and $B N \perp XY$
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To prove :
(i) $\triangle O A M \cong \triangle O B N$
(ii) $A M=B N$
Proof :
(i)In \triangle OAM and \triangle OBN
$O A=O B$
$\angle M=\angle N$
and $\angle A O M=\angle B O N$
So $\triangle O A M \cong \triangle O B N$
Question 6
Sol: In the figure '
If YP is the bisector of $\angle xyz$
L is any point on YB and MLN is perpendicular to YP
To prove: $L M=L N$
Proof : $\triangle YML$ and $\triangle YN L$
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YL=YL
$\angle Y L M=\angle YL N$
$\angle M Y L=\angle N Y L$
So $\triangle YM L \cong \triangle Y N L$
So $L M=L N$ proved
Question 7
Sol:
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In the figure
ABCD and AXYZ are square where X is any point within the square
ABCD BX and DZ are joined
To prove :BX = DZ
Proof :In $\triangle A B X and $\triangle$ A O Z
$A B=A D$
$A X=A Z$
$\angle B A X=\angle D A Z$
So $\triangle A B X \cong \triangle A D Z$
So $B X=D Z$
Proved
Question 8
Sol: (Image to be added)
In the figure . P is any point inside the $\triangle BAC$
$P M \perp A B$ and PN \perp A$ are drawn and PM=PN
To prove : AP is bisector of $\angle B A C$
Proof : In right $\triangle$ PAM AND $\triangle$ PAN
PM=PN
PA = PA
So $\triangle P A M \cong \triangle P A N$
$\angle P A M=\angle P A N$
A is the bisector of $\angle B A C$
proved
Question 9
Sol: In the figure PR=PS,PT=PQ $\angle P QR$=$90^{\circ}$ and $\triangle P T S=90^{\circ}$
To prove:
(i) RQ= ST
(ii) RT=SQ
Proof:
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(i)$\Delta P Q R$ and $\triangle PTS$ is
$\begin{aligned}&P Q=P T \\&P R=P S\end{aligned}$
So $\triangle P Q R \cong \triangle P T S$
So $R Q=S T$
(ii)Is PR=PS
PT=PQ
Subtracting we get :
PR-PT =PS -PQ
=RT= SQ
Hence proved
Question 10
Given : In square PQRS are AB is drawn with center
P and a radius less than PR which intersects SR at A and
RQ at B
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To prove :AS = BQ
Construction : Join PA and PB
Proof : In right $\triangle P Q B$ and $\triangle P S A$
$P Q=P S$
$P B=P A$
SO $\triangle P Q B \cong \triangle P S A$
So $\begin{aligned} B Q &=A S \\ \text { or } A S &=B Q \end{aligned}$
Proved
Question 11
Sol: In the figure, $\triangle P Q R$ in which $L M=M N, QM=MR$ and
$\angle M L Q=\angle M N R=90^{\circ}$
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To prove: $P Q=P R$
Proof : In right $\triangle Q L M$ and $\triangle R N M$
LM = MN
QM = MR
So $\triangle Q L M \cong \triangle R N M$
So $\angle Q=\angle R$
Now in $\triangle P Q R$
$\begin{aligned} \angle Q &=\angle R \\ P R &=P Q \end{aligned}$
or PQ=PR
Proved
Question 12
Sol: Given $\triangle A B C$ is a right angled, right angle at B. ACDE and
BCGF are squares on the sides AC and BC respectively
AG and BD are joined
To prove :
(i) $\triangle B C D \cong \triangle A C G$
(ii)$A G=B D$
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Proof: if $\angle A C D=\angle B C A$
Adding $\angle A C D$ to both sides,
$\angle A C D+\angle A C B=\angle A C B+\angle B C G$
$\Rightarrow \angle B D C=\angle A C G$
Now in $\triangle B C D$ and $\triangle A C G$
BC=CG
$\angle B C D=\angle A C G$
$C D=A C$
$\triangle B C D \cong \triangle A C G$
$B D=A G$
or $A G=B D$
Proved'
Question 13
Sol: $\triangle A B C$ is an equilateral triangle in which AD, BE and CF Are its medians:
To prove : AD=BE=CF
Proof: $\triangle EBC and \triangle FBC$
BC = BC (Common)
EC = FB
$\angle C=\angle B$
So $\triangle E B C \cong \triangle F B C$
So BE =CF .................(i)
Similarly we can prove that
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$\triangle E B C \cong \triangle A D C$
$B E=A D$.......(ii)
From (i) and (ii)
AD=BE = CF
Proved
Question 14
Sol: In the figure
AC=DE, $\angle A C B=\angle E D F$ and $B C=F C$
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To prove :AB=EF
Proof : BD=CF
Adding DC to both sides
BD+DC+=DC+CF
BC=DF
Now in $\triangle A B C$ and $\triangle D E F$
$A C=D E$
$B C=D E$
$\angle A C B=\angle E D F$
So $\triangle A B C \cong \triangle D E F$
So $A B=E F$
proved
Question 15
Sol: In the figures
AC= AE, AB=AD and $\angle B A D=\angle E A C$
TO prove : BC=DE
Construction : Join DE
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Proof :$\angle B A D=\angle E A C$
So $\angle B A D+\angle D A C=\angle D A C+\angle E A C$
In $\triangle A B C$ and $\triangle A D E$
$A B=A D$
$A C=A E$
$\angle B A C=\angle D A E$
So $\triangle A B C \cong \triangle A D E$
So BC=DE Proved
Question 16
Sol: In $\triangle P Q R$ PS is the median of side QR QL AND RM are perpendicular on the median
To prove: QL=RM
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Proof : In $\triangle Q L S$ and $\triangle R MS $
$Q S=S R$
$\angle Q S L=\angle R S M$
$\angle Q L S=\angle R M S$
$\triangle Q L S \cong \Delta R M S$
$Q L=R M$
Proved
Question 17
Sol: ABCD is a parallelogram sides AB and AD are produced to E and F respectively such that AB = BE And AD = DF
To prove : $\triangle B E C \cong \triangle D C F$
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Construction : Join EC and FC
Proof : In parallelogram ABCD
$\angle D A B=\angle C D C=\angle C B E$
Now In $\triangle B E C$ and $\triangle D C F$
$B C=D F$
$B F=D C$
and $\angle C B E=\angle F D C$
So $\triangle B E C \cong \triangle D C F$ Proved
Question 18
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Sol: The figure , QX and RX are the bisectors of the angles Q and R respectively of the triangle PQR AND $XS\perp QR$ and $X T \perp P Q$
To prove:
(i)$\triangle X T Z \cong \triangle XS Q$
(ii) PX Bisects the angle P
Construction : Join PX and draw $XV \perp P R$
Proof : $\triangle X Q S$ and $\triangle X T Q$
$XQ=\times Q$
$\angle S=\angle T$
$\angle X Q S=\angle XQ T$
$\triangle XQ S \cong \triangle X T Q$
So XS =XT
Similarly we can prove that
$\triangle X S R \cong \triangle X V R$
So XS=XV .........(ii)
From (i) and (ii) XT=XV
Now in right triangle XTP and $\triangle X Y P$
$X P=X P$(Common)
$X T=X V$
So XTP$\cong \triangle X Y P$
$\angle X P T=\angle X P V$
P X bisects $\angle P$
Proved
Question 19
Sol: (i) Figure (i)$A B \| D E$ And BC =CD in $\triangle A B C$ and $\triangle D E C$.
$\begin{aligned} B C &=C D \\ \angle A C B &=\angle F C D \\ \angle B &=\angle D . \end{aligned}$
SO $\triangle A B C \cong \triangle D E C$
SO $A B=D E$
=$2 x-4=14$
=$2 x=14+4=18$
$x=\frac{18}{2}=9$
$x=9$
and $A C=C E$
$\Rightarrow 3 y+5=20$
$\Rightarrow 3 y=20-5=15$
$y=\frac{15}{3}=5$
Hence $x=9$ and $y=5$
(ii) In figure (ii)
$A B=A D, B C=D C$
$\angle B A C=(y-6)^{\circ}, \angle B C A=63^{\circ}$
$\angle C A D=30^{\circ}$ and $\angle A C D=(2 x+7)$
Now $\triangle A B C$ and $\triangle A D C$
$A C=A C$
$A B=A D$
$B C=D C$
So $\triangle A B C \cong \triangle A D C$
So $\angle B A C=\angle C A D$
So $\angle B A C=30^{\circ}$
$y-6^{\circ}=30^{\circ}$
$y=30^{\circ}+6^{\circ}=36^{\circ}$
and $\angle B C A=\angle A C D$
$63^{\circ}=2 x+7^{\circ}$
$\Rightarrow 2 x=63-7=56^{\circ}$
$\Rightarrow x=\frac{56^{\circ}}{2}=28^{\circ}$
Hence $x=28^{\circ}$ and $y=36^{\circ}$
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