Solutions:

1.   In Fig. 6.17, (i) and (ii), DE || BC. Find EC in (i) and AD in (ii).


Solution 
In ∆ABC, DE || BC. Therefore, by theorem ,
 If a line is drawn parallel to one side of a triangle to intersect the other two sides in 
 distinct points, the other two sides are divided in the same ratio.
 
 i.e ${AD}/{DB}={AE}/{EC}$
 i.e ${1.5}/{3}={1}/{EC}$
 i.e ${1.5}/{3}={1}/{EC}$
 $EC=2$ cm.
 
 In fig (ii),
 i.e ${AD}/{DB}={AE}/{EC}$
 i.e ${AD}/{7.2}={1.8}/{5.4}$
 i.e ${AD}={7.2}/{3}$=$2.4$ cm
 
 2.   E  and  F  are  points  on  the  sides  PQ  and  PR respectively of a ∆PQR. For each of the following cases, 
state whether EF || QR :
(i)   PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm

Solution 
In ∆PQR, if EF || QR , then
		${PE}/{EQ}={PF}/{FR}$
		So,${PE}/{EQ}=3.9/3 = 1.3$
		${PF}/{FR}=3.6/2.4=3/2=1.5$
		
	Since, ${PE}/{EQ}≠{PF}/{FR}$, therefore  EF is not parallel to QR.
.

(ii)   PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm
Solution 
In ∆PQR, if EF || QR , then ${PE}/{EQ}={PF}/{FR}$
		So,${PE}/{EQ}= 4/4.5=8/9$
		and ${PF}/{FR}= 8/9$
	Since ${PE}/{EQ}={PF}/{FR}$, EF is parallel to QR . 
		
(iii)   PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.36 cm

Solution 
In ∆PQR, if EF || QR , then ${PE}/{EQ}={PF}/{FR}$
		So,${PE}/{EQ}= 1.28/2.56=1/2$
		and ${PF}/{FR}= 0.18/0.36=1/2$
		Since ${PE}/{EQ}={PF}/{FR}$, EF is parallel to QR . 

3.   In Fig., if  LM || CB and LN || CD, prove that

    ${AM}/{AB}={AN}/{AD}$
	
	Solution:
	
	 In ∆ABC, since LM || BC, 
	         ${AM}/{MB}={AL}/{LC}$ or ${MB}/{AM}= {LC}/ {AL}$
		Adding 1 on both side,
	         ${MB}/{AM}+1={LC}/ {AL} +1$
			 
	         ${MB+AM}/{AM }={LC+AL}/{AL}$
			 Since (MB+AM)=AB and LC+AL = AC
	         ${AB}/{AM }={AC}/{AL}$  ..............(1)
			 
	Similairly in ∆ACD, since LN||CD,
	        ${AN}/{ND}={AL}/{LC}$ or  ${ND}/{AN}={LC}/{AL}$
			
			Adding 1 on both sides,
			${ND}/{AN}+1={LC}/{AL}+1$
			
			${AN+ND}/{AN}={AL+LC}/{AL}$
			Since AN+ND = AD and AL+LC=AC
			${AD}/{AN}={AC}/{AL}$ ...........(2)
			
	Therefore from eqn (1)  and  (2)
			${AB}/{AM }={AC}/{AL}= {AD}/{AN}$
			
		i.e ${AB}/{AM }= {AD}/{AN}$
		
		i.e ${AM}/{AB}={AN}/{AD}$.. Hence Proved. 
	      
				
4.   In Fig. DE || AC and DF || AE. Prove that
     ${BF}/{FE}={BE}/{EC}$
	 
	 Solution:
	 
	 
	 In ∆ABC,since DE|| AC,
	 
	 ${BD}/{DA}={BE}/{EC}$ ...................(1)
	 
	 In ∆ABE, DF||AE,
	 
	 Therefore, ${BD}/{DA}={BF}/{FE}$ ..............(2)

	From (1) and (2), ${BF}/{FE}={BE}/{EC}$  ..Hence Proved  
	 

5.   In Fig. 6.20, DE || OQ and DF || OR. Show that
     EF || QR.
	 
	 Solution:
	 

	 In ∆OPQ, since DE || OQ,
	 
	 ${PE}/{EQ}={PD}/{DO}$ ............(1)
	 
	 In ∆OPR, since DF || OR,
	 
	 ${PF}/{FR}={PD}/{DO}$  .............(2)
	 
	 Therefore from (1) and (2) ,
	 
	 ${PE}/{EQ}= {PF}/{FR}$ 
	 
	 Since E & F are the points on PQ & PR of ∆PQR, and if ${PE}/{EQ}= {PF}/{FR}$ 
	 then EF must be parallel to QR. 
	  Hence Proved..
	 
6.   In Fig. 6.21, A, B and C are points on OP, OQ and OR respectively such that AB || PQ and AC || PR.
	 Show that BC || QR.
	 
	 Solution:
	 	 
	 
	 In ∆OPQ, since AB || PQ,
	 
	 Therefore, ${OA}/{AP}= {OB}/{BQ}$ ...........(1)
	 
	 In ∆OPR, since AC || PR,
	 
	 Therefore, ${OA}/{AP}={OC}/{CR}$  ...........(2)
	 
	 From (1) and (2), ${OB}/{BQ}= {OC}/{CR}$ ..........(3)
	 
	 Since B & C are the points on OQ & OR of ∆OQR, therefore BC must be parallel to
	 
	 QR to satisfy the above relation.
	 
7.   Using Theorem 6.1, prove that a line drawn through the mid-point of one side of a triangle parallel
     to another side bisects the third side.
	 
	 Solution:
	 
	 Since according to theorem,
	  If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points,
	 the other two sides are divided in the same ratio.
	 
	 Therefore in ∆ABC,since  DE || BC
	 
	 ${AD}/{DB}={AE}/{EC}$
	 
	 Since D is mid point of AB, AD= DB
	 
	 ${AD}/{AD}={AE}/{EC}$
	 
	 ${EC}={AE}$
	 
	 Since E is a point on AC and AE=EC, so E is the mid point of AC. 
	 	 
8.   Using Theorem 6.2, prove that the line joining the mid-points of any two sides of a triangle is 
      parallel to the third side.
	  
	  Solution:
	   
	  In ∆ABC, let D be the mid point on AB and E be the mid point on AC.
	  
	  Therfore $AD=DB$ and $AE=EC$
	  
	  Ratio of ${AD}/{DB}=1$ and ${AE}/{EC}=1$
	  
	  Therefore ${AD}/{DB}= {AE}/{EC}$
	  
	  This relationship is only fulfilled be DE || BC. Hence Proved.
	  
9.   ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O. 
     Show that ${AO}/{BO}={CO}/{DO}$
	 
	 Solution:
	 
	 	 
	 Let draw a line passing through O, parallel to AB and intersecting at E on AD and F on BC.
	 
	 Therefore in ∆ACD, ${AE}/{ED}={AO}/{OC}$ ............(1)
	 
	 ∆BCD, ${BF}/{FC}={BO}/{OD}$   .................(2)
	 
	 ∆ABD, ${AE}/{ED}={BO}/{OD}$  .................(3)
	 
	 ∆ABC, ${AO}/{OC}={BF}/{FC}$  ..................(4)
	 
	 Therefore from (1),(2), (3) and  (4)
	 
	 ${AO}/{BO}={CO}/{DO}$  Hence Proved .
	 
	 
10. The diagonals of a quadrilateral ABCD intersect each other at the point O such that
    ${AO}/{BO}={CO}/{DO}$.Show that ABCD is a trapezium.