Exercise 9 A
Question 1
Sol : (i) The line joining the mid point of two sides of a triangle is parallel to the third side.
(ii) The line drawn through the mid point of one side of a triangle parallel to another side bisects the third side.
(iii) In the figure S and T are the mid point of PQ and PR respectively .If ST=3cm then QR=6cm
(iv) In figure ' D and E are the mid point of AB and AC.
If DE = 7.5cm then BC=15cm
(v) (image to be added)
(a) In the given figure
In $\triangle A B C$
$A D=D B \text { and } C E=E B$
So D and E are mid points of AB and BC Respectively
So $DE\| AC$
So $D E=\frac{1}{2} A C$ and $D E \| A C$
$\begin{aligned} 64 &=\frac{1}{2} \times 4 x \\ \Rightarrow 2 x &=64 \Rightarrow x=\frac{64}{2}=32 \end{aligned}$
(b) (image to be added)
In AB=DB and AE=EC
So $DE\| B C$
$D E=\frac{1}{2} B C$
$\Rightarrow x-8=\frac{1}{2} \times 40 \Rightarrow x-8=20$
$\Rightarrow x=20+8=28$
So $x=28$
(C) In $\triangle A B C$
$A D=D C$ and $B E=E C$
SO $D E \| A B$ and $D E=\frac{1}{2} A B$
$x+8=\frac{1}{2} x 6 x \Rightarrow x+8=3 x \Rightarrow 3 x-x=8$
$\Rightarrow 2 x=8 \Rightarrow x=\frac{8}{2}=4$
So $x=4$
Question 2
(Image to be added)
Sol: In $\triangle A B C, A B=6 \mathrm{~cm}, A C=3 \mathrm{~cm} \mathrm{~M}$ is mid point of AB
A line through M is drawn parallel to AB which cuts BC in N If m IS MID point of AB and MN$\|$AC
So $M N=\frac{1}{2} A C$
$=\frac{1}{2} \times 3 \mathrm{~cm}=1.5 \mathrm{~cm}$
Question 3
(IMAGE TO BE ADDED)
Sol: In the figure
$\triangle A B C, D$ is mid point of AB and E is mid point of AC
So $D E \| B C$ and $D E=\frac{1}{2} B C$
(i) If BC = 6cm
So $D E=\frac{1}{2} \times 6=3 \mathrm{~cm}$
(ii)if $\angle B B C=140^{\circ}$
if $D E\|B C$
So $\angle A D E=\angle D B C$ or $\angle A B C$
So ADE = $140^{\circ}$
Question 4
(IMAGE TO BE ADDED)
Sol: D,E and F are the mid points of the sides BC, CA and AB Respectively AD and EF are joined
To prove: AD bisects EF
Construction: Join ED and FD
Proof: In $\triangle A B C$
E and F are the mid points of CA and AB
So, Ef $\| B C$ and $E F=\frac{1}{2} B C$
Similarly we can prove that
ED$\|$AB and equal to $\frac{1}{2} A B$ and FD|| AC and equal to $\frac{1}{2} A C$
So AEDF is a parallelogram
If the diagonals bisect each other
So AD bisects EF
PROVED
Question 5
Sol: $\triangle A B C$ is an isosceles triangle in which AB=AC
D,F and F are the mid points of BC, AB and AC respectively
AD and EF are joined
To prove: AD and EF are perpendicular to each other
Proof : If E and F are mid points of AB and AC
So, AE =EB and AF =FC
But AB=AC
So AE =AF
and $E F \| B C$ and $E F=\frac{1}{2} B C$
Now in $\triangle A B D$ and $\triangle A C D$
$A B=A C$
$B D=D C$
$A D=A D$
So $\triangle A B D \cong \triangle A C D$
So $\angle B A D=\angle C A D$
Now in $\triangle A E G$ and $\triangle A F C$
$A G=A G$ (Common)
$A E=A F$
$\angle B A D$ or $\angle E A G=\angle C A D or \angle A F G$ Proved
So
$\begin{aligned} & \triangle A E G \cong \triangle A F G \\ \text { So } \angle A G E &=\angle A G F \end{aligned}$
But $\angle A G E+\angle A F G=180^{\circ}$
$\text { So } \angle A G E=\angle A G F=90^{\circ}$
So AD and EF are perpendicular to each other
Hence proved
Question 6
Sol: In $\triangle A B C, D$ and F are the mid point of the sides BC, CA and AB respectively
DE , $E F$ and $F D$ are joined
To prove: $\triangle A E F \cong \triangle B D F$
$\triangle D E F \cong \triangle C D E$
Proof : if D and E are the mid points of side BC and CA respectively
So $D E \| A B$ and $D E=\frac{1}{2} A B$
So $D E=B F=F A$ .........(i)
Similarly , D and F are the mid point of BC and AB respectively
So DF $\| A C$ and $D F=\frac{1}{2} A C$ .............(ii)
$DF=A E=E C$
Now in $\triangle A E F$ and $\triangle D E F$
$A F=D E \quad$ proved
$A E=D F \quad$ proved
$E F=E F \quad $ (common)
So $\triangle A E F \cong \triangle D E F$ ............(i)
Similarly we can prove that
$\triangle B D F \cong \triangle D E F$ .............(i)
and $\triangle C D E \cong \triangle D E F$......(ii)
From (i),(ii) and (iii)
$\triangle A E F \cong \triangle B D F \cong \triangle C D E \cong \triangle D E F$ Hence proved
Question 7
(IMAGE TO BE ADDED)
Sol: $\triangle A B C$ is an equilateral triangle D, E and F are the Mid point of sides BC, CA and AB respectively DE, EF and FD are joined
To prove : $\triangle D E F$ is an equilateral triangle
Proof: If E and F are the mid points of AC and AB respectively
So $E F \| B C$ and $E F=\frac{1}{2} B C$...........(i)
Similarly , D and E are then mid points to BC and AC Recpectively
So $\therefore \quad D E \quad \| A B$ and $D E=\frac{1}{2} A B$
and D and F are the mid points of BC and AB respectively
So $D F \| A C$ and $D F=\frac{1}{2} A C$...........(iii)
from (i) (ii) and (iii)
if $A B=B C=C A$
So $\frac{1}{2} E F=\frac{1}{2} D E=\frac{1}{2} D F$
$\Rightarrow D E=E F=F D$
So $\triangle D E F$ is an equilateral triangle
Hence proved
Question 8
(IMAGE TO BE ADDED)
Sol: In $\triangle A B C, A D$ and BE are tits medians $B E \| D F$
To prove : C F=$\frac{1}{4} A C$
proof : IF BE is the median
So E is mid point of AC
So $A E=E C$ or $E C=\frac{1}{2} A C$..........(i)
In $\triangle B C E$,
D is mid point of BC and $D F \| B E$
So F is mid point of EC
So $F C=\frac{1}{2} E C$ ..............(iv)
From (i) and (ii)
$C F=\frac{1}{2} E C=\frac{1}{2}\left(\frac{1}{2} A C\right)=\frac{1}{4} A C$
Hence $C F=\frac{1}{4} A C$
Question 9
Sol: $\triangle A B C$ is an isosceles triangle in which AB=AC
CP||BA is drawn and AP is the bisector of $\angle C A D$
(IMAGE TO BE ADDED)
TO Prove:
(i) $\angle P A C=\angle B C A$
(ii)BCP is a parallelogram
Proof: In ABC
(i) If AB=AC
So $\angle C=\angle B$
But Ext. $\angle C A D=\angle B+\angle C=2 \angle C$.....(i)
and $\angle P A C=\angle P A D=\frac{1}{2} \angle C A P$.............(ii)
So from (i) and (ii)
$\angle C=\angle C A P$ or $\angle P A C=\angle B C A$
(ii)So $\angle P A C=\angle B C A$
But these are alternate angles
So AP||BC
But CP||BA
So ABCP is a parallelogram Hence proved
Question 10
Sol: In the figure , ABCD is a kite in which AB=AD and BC = DC
P,Q,R and S are the mid points of the sides AB,AD , DC
and BC respectively PQ, QR, RS and PS are joined
To prove: PQRS is a rectangle
Construction: Join BD and AC
Proof: In $\triangle A B D$
If P and Q are mid point of AB and AD respectively
So PQ||BD and $P Q=\frac{1}{2} B D$...........(i)
Similarly in $\triangle B C D$,
S and R are the mid points of BC and DC
So $S R \| B D$ and $S R=\frac{1}{2} B D$ ...........(ii)
From (i) and (ii)
So PQ=SR and PQ ||QR
So PQRS is a parallelogram
If diagonals AC and BD intersect
Each other at right angles
So PQRS is a rectangle
Hence proved
Question 11
Sol: ABCD is a rectangle P,Q,R and S are the mid point of the sides AB, BC , CD and DA respectively PQ, QR, RS and SP are joined
TO prove: PQRS is a rhombus
Construction: Join PR and SQ, AC and BD are also joined
Proof: In $\triangle A B C$
If P and Q are mid points of AB and BC respectively
(IMAGE TO BE ADDED)
So PQ||AC and $P Q=\frac{1}{2} A C$ ............(i)
Similarly in $\triangle A D C$
S and R are the mid points of AD and CD respectively
So SR||AC and $S R=\frac{1}{2} A C$............(ii)
From (i) and (ii)
PQRS is a parallelogram
Similarly PS||BD and SP= $=\frac{1}{2} \quad BD$
and QR||BD and QR =$\frac{1}{2} B D$
But in rectangle ABCD diagonals AC and BD are equal
PQRS is a rhombus
Hence proved
Question 12
Sol: ABCD is a quad which is a rhombus P,Q,R and S are the mid points of the side AB, BC, CD and DA Respectively PQ, QR, RS and SP are joined
(IMAGE TO BE ADDED)
To prove: PQRS is a rectangle
Construction: Join AC and BD
Proof : In $\triangle A B C$
If P and Q are the mid point of AB and CD respectively
So $P Q \| A C$ and $P Q=\frac{1}{2} A C$ ............(i)
Similarly in $\triangle A D C$
R and S are the mid points of CD and DA respectively
So RS||AC and RS$=\frac{1}{2} AC$...........(ii)
So PQRS is a parallelogram
If diagonals as rhombus bisect each other at right angles PQRS is a rectangle Hence proved
Question 13
Sol: In quadrilateral ABCD L,M,P and Q are the mid points of the sides AB, BC , CD and DA respectively . LM and AC are joined
(IMAGE TO BE ADDED)
To prove:
(i) Using the mid point theorem make a statement concerning the lengths and direction of LM and AC
(ii) Prove that LMPQ is a parallelogram .
(iii) If its is also given that diagonals AC and BD are equal what further statement can be made about the parallelogram LMPQ
Construction: Join BD, MP , PQ and QL
Proof:
(i) if Land $m$ are mid-points of $A B$ and $B C$ respectively.
So $L m \| A C$ and $L m=\frac{1}{2} A C$.........(i)
(ii)Similarly P and Q are the mid points of CD and DA respectively
So $P Q \| A C$ and $P Q=\frac{1}{2} A C$.............(ii)
From (i) and (ii)
LMPQ is a paralleogram
(iii) If diagonals AC and BD are equal
Then ABCD can be a parallelogram or a rectangle
Then LM=MP =PQ=QL
Then LMPQ will be rhombus
(If is the sides of a parallelogram are equal then it is a rhombus )
Question 14
Sol: In the figure ABCD is a parallelogram, E and F are the mid points of AB and CD respectively GH is any line which Intersect AD in G. EF in P and BC in H
To prove: GP =PH
Proof: If E and F are the mid points of AB and DC respectively
So EF||AD||BC (If ABCD is a parallelogram)
So AE=EB = $\frac{A E}{E B}=1$
So $\frac{G P}{P H}=1 \Rightarrow G P=P H$ (intersect theorem)
Hence proved
Question 15
Sol: In ||gm ABCD , E and F are the mid points of side AD and BC respectively
EB and DF are joined which intersect diagonal AC at M and N
To prove: M and N trisect AC
i.e AM=MN=NC
Proof: If E and F are the mid points of AD and BC respectively
So BE||DE
Now in $\triangle$ ADN
if CF=FB
So CN= NM
from (i) and (ii)
AM=MN= NC
Hence M and N trisect AC
Question 16
Sol: In trapezium ABCD , AB||DC E is mid point of AD
A line through E is drawn parallel to AB which meets BC in F
To prove : F is the mid point to BC
Proof: IN $\triangle A D B$
In E its mid point of AD and ED OR EF is parallel to AC
So
O is mid point of BD
Similarly in $\triangle B C D$
O is mid point of BD and E OF ||CD
So F is mid point BC
Hence proved
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