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TOPICS
Exercise - 6.2

Question-1 :-  In Figure, (i) and (ii), DE || BC. Find EC in (i) and AD in (ii).

Solution :-
```(i) Given that DE || BC.
Let EC = x cm
By using basic proportionality theorem, we get
1.5/3 = 1/x
1.5x = 3
x = 3/1.5
x = 30/15
x = 2
Hence, EC = x cm = 2 cm
```
```(ii) Given that DE || BC.
By using basic proportionality theorem, we get
x/7.2 = 1.8/5.4
5.4x = 1.8 x 7.2
5.4x = 12.96
x = 12.96/5.4
x = 2.4
Hence, AD = x cm = 2.4 cm
```

Question-2 :-  E and F are points on the sides PQ and PR respectively of a Δ 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
(ii) PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm
(iii) PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.36 cm

Solution :-
```(i) Given that PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm, FR = 2.4 cm

PE/EQ = 3.9/3 = 1.3
PF/FR = 3.6/2.4 = 1.5
Hence, PE/EQ ≠ PF/FR
Therefore, EF is not parallel to QR.
```
```(ii) Given that PE = 4 cm, QE = 4.5 cm, PF = 8 cm, RF = 9 cm

PE/EQ = 4/4.5 = 8/9
PF/FR = 8/9
Hence, PE/EQ = PF/FR
Therefore, EF || QR.
```
```(iii) Given that PE = 4 cm, QE = 4.5 cm, PF = 8 cm, RF = 9 cm

PE/PQ = 0.18/1.28 = 18/128 = 9/64
PF/PR = 0.36/2.56 = 9/64
Hence, PE/PQ = PF/PR
Therefore, EF || QR.
```

Question-3 :-  In Figure, if LM || CB and LN || CD, prove that AM/AB = AN/AD

Solution :-
```  Given that : LM || CB and LN || CD
Prove that : AM/AB = AN/AD
Proof : LM || CB                      (Given)
By using basic proportionality theorem, we get
AM/AB = AL/AC .........(i)
Similarly, LN || CD
From (i) & (ii), We get
```

Question-4 :-  In Figure, DE || AC and DF || AE. Prove that BF/FE = BE/EC

Solution :-
```  Given that : DE || AC and DF || AE
Prove that : BF/FE = BE/EC
Proof : In ΔABC, DE || AC                   (Given)
By using basic proportionality theorem, we get
BD/DA = BE/EC .........(i)
Similarly, In ΔBAE, DF || AE
BD/DA = BF/FE .........(ii)
From (i) & (ii), We get
BE/EC = BF/FE
```

Question-5 :-  In Figure, DE || OQ and DF || OR. Show that EF || QR.

Solution :-
```  Given that : DE || OQ and DF || OR
Prove that : EF || QR
Proof : In Δ POQ, DE || OQ                   (Given)
By using basic proportionality theorem, we get
PE/EQ = PD/DO .........(i)
Similarly, In Δ POR, DF || OR
PF/FR = PD/DO .........(ii)
From (i) & (ii), We get
PE/EQ = PF/FR
By using converse of basic proportionality theorem
Therefore, EF || QR
```

Question-6 :-  In Figure, A, B and C are points on OP, OQ and OR respectively such that AB || PQ and AC || PR. Show that BC || QR.

Solution :-
```  Given that : AB || PQ and AC || PR
Prove that : BC || QR
Proof : In Δ POQ, AB || PQ                   (Given)
By using basic proportionality theorem, we get
OA/AP = OB/BQ .........(i)
Similarly, In Δ POR, AC || PR
OA/AP = OC/CR .........(ii)
From (i) & (ii), We get
OB/BQ = OC/CR
By using converse of basic proportionality theorem
Therefore, BC || QR
```

Question-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. (Recall that you have proved it in Class IX).

Solution :-
```  Given that : Consider the given figure in which PQ is a line segment drawn through the mid-point P of line AB, such that PQ || BC

Prove that : Q is the mid-point of AC.
Proof : In Δ ABC, PQ || BC
By using basic proportionality theorem, we get
AQ/QC = AP/PB
AQ/QC = 1/1       (P is the mid-point of AB. So, AB = PB)
AQ = QC
or Q is the mid point of AC.
```

Question-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. (Recall that you have done it in Class IX).

Solution :-
```  Given that : Consider the given figure in which PQ is a line segment joining the mid-points P and Q of line AB and AC respectively.
So, we have AP = PB and AQ = QC.

Prove that : PQ || BC
Proof : In Δ ABC
AP/PB = 1/1       (Given AP = PB)
AQ/QC = 1/1       (Given AQ = QC)
Therefore, AQ/QC = AP/PB
By using basic proportionality theorem, we get
PQ || BC
```

Question-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 :-
```  Given that : ABCD is a trapezium in which AB || DC.
Prove that : AO/BO = CO/DO
Construction : Draw a line EF through point O, such that EF || CD

Proof : In Δ ADC, EO || CD
By using basic proportionality theorem, we get
AE/ED = OA/OC  ........(i)
In Δ ABD, OE || AB
By using basic proportionality theorem, we get
ED/AE = OD/OB
AE/ED = OB/OD  ........(ii)
From equations (i) & (ii), we get
OA/OC = OB/OD
Therefore, AO/BO = OC/OD
```

Question-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.

Solution :-
```  Given that : The diagonals of a quadrilateral ABCD intersect each other at the point O such that AO/BO = CO/DO.
Prove that : ABCD is a trapezium.
Construction : Draw a line OE || AB.

Proof : In Δ ABD, OE || AB
By using basic proportionality theorem, we get
AE/ED = OB/OD  ........(i)
AO/BO = CO/DO  ........(ii) (Given)
From equations (i) & (ii), we get
AE/ED = AO/OC
EO || DC                    (By the converse of basic proportionality theorem)
AB || OE || DC
AB || CD
Therefore, ABCD is a trapezium.
```
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