SChand CLASS 9 Chapter 9 Mid Point and Intercept Theorems EXERCISE 9(A)

  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