Triangles
Question 1.
In figures, (i) and (ii), DE || BC. Find EC in (i) and AD in (ii).
Solution:
(i) In Fig. (i),
since DE || BC, therefore
AD/DB = AE/EC
or 1.5/3 = 1/EC
or EC = 3/1.5
or EC = (3×10)/15 cm
∴ EC = 2 cm
(ii) In Fig. (ii),
since DE || BC, therefore
AD/DB =AE/EC
or AD/7.2 = 1.8/5.4
or AD = 18/54 x 72/10 cm
∴ AD = 2.4 cm.
Question 2.
E and F are points on the sides PQ and PR respectively of a 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
(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) We have :
PE = 3.9 cm, EQ = 3 cm,
PF = 3.6 cm and FR = 2.4 cm
i. e., EF does not divide the sides PQ and PR of A PQR in the same ratio. Therefore, EF is not parallel to QR.
(ii) We have : PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm
Now, PE/EQ = 4/4.5
or PE/EQ = 40/45
∴ PE/EQ = 8/9
and PF/FR = 8/9
So, PE/EQ = PF/FR
Thus, EF divides sides PQ and PR of ∆ PQR in the same ratio. Therefore, by the converse of Basic Proportionality Theorem, we have EF || QR.
(iii) We have :PQ= 1.28 cm, PR = 2.56 cm,
PE = 0.18 cm and PF 0.36 cm
or EQ = PQ – PE
or EQ = (1.28 – 0.18) cm
∴ EQ = 1.10 cm
and FR = PR – PF
or FR = (2.56 – 0.36)
∴ FR = 2.20 cm
Now, PE/EQ = 0.18/1.10
or PE/EQ = 18/110
∴ PE/EQ = 9/55
and PF/FR = 0.36/2.20
or PF/FR = 36/220
∴ PF/FR = 9/55
So, PE/EQ = PF/FR
Thus, EF divides sides PQ and PR of ∆ PQR in the same ratio. Therefore, by the converse of Basic Proportionality Theorem, we have EF || QR.
Question 3.
In the figure, if LM || CB and LN || CD, prove that AM/AB = AN/AD
Solution:
In ∆ ABC, we have :
LM || CB
∴By a result based on Basic Proportionality Theorem (a corollary), we have :
AM/AB = AL/AC … (1)
In ∆ ACD, we have :
LN || CD [Given]
∴By a result based on Basic Proportionality Theorem, we have :
AL/AC = AN/AD
From (1) and (2), we obtain that
AM/AB = AN/AD
Question 4.
In the figure, DE || AC and DF || AE. Prove that BF/FE = BE/EC
Solution:
In ∆ BGA, we have :
DE || AC
∴By Basic Proportionality Theorem, we have :
BE/EC = BD/DA …. (1)
In ∆ BEA, we have :
DF || AE [Given]
∴By Basic Proportionality Theorem, we have
BF/FE = BD/DA …. (2)
From (1) and (2), we obtain that
BF/FE = BE/EC.
Question 5.
In the figure, DE || OQ and DF || OR. Show that EF || QR.
Solution:
In ∆ PQO, we have :
DE || OQ [Given]
∴By Basic Proportionality Theorem, we have :
PE/EQ = PD/DO … (1)
In ∆ POR, We have :
DF || OR [Given]
∴By Basic Proportionality Theorem, we have
PD/DO = PF/FR … (2)
From (1) and (2), we obtain that
PE/EQ = PF/FR
So, EF || QR [By the converse of BPT]
Question 6.
In the 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 : O is any point within ∆ PQR, AB || PQ and AC || PR
To prove : BC || QR.
Construction : Join BC.
Proof:
In ∆ OPQ, we have :
AB || PQ [Given]
∴By Basic Proportionality Theorem, we have :
OA/AP = OB/BQ … (1)
In ∆ OPR, We have :
AC || PR [Given]
∴ By Basic Proportionality Theorem, we have
OA/AP = OC/CR … (2)
From (1) and (2), we obtain that
OB/BQ = OC/CR
Thus, in ∆ OQR, B and C are points dividing the sides OQ and OR in the same ratio. Therefore, by the converse of Basic Proportionality Theorem, we have :
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 : ∆ ABC, in which D is the mid-point of side AB and the line DE is drawn parallel to BC, meeting AC in E.
To prove : AE = EC
Proof:
Since DE || BC, therefore by Basic Proportionality Theorem, we have :
AD/DB = AE/EC … (1)
But AD = DB [∵D is the mid-point of AB]
i.e., AD/DB = 1
∴ From (1), AE/EC = 1
or AE = EC
Hence, E is the mid-point of the third side 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 : ∆ ABC, in which D and E are the mid¬points of sides AB and AC respectively.
To prove : DE || BC.
Proof :
Since D and E are the mid-points of AB and AC respectively, therefore
AD = DB and AE = EC
Now, AD = DB
∴ AD/DB = 1
and AE = EC
∴ AE/EC = 1
Thus, in ∆ ABC, D and E are points dividing the sides AB and AC in the same ratio. Therefore, by the converse of Basic Proportionality Theorem (Theorem 6.2), we have :
DE || 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 : A trapezium ABCD, in which AB || DC and its diagonals AC and BD intersect each other at O.
To prove : AO/BO = CO/DO.
Construction : Through O, draw OE || AB, i.e., OE || DC.
Proof:
In ∆ ADC, we have :
OE || DC [Construction]
∴ By Basic Proportionality Theorem, we have :
AE/ED = AO/CO … (1)
Again, in ∆ ABD, we have :
OE || AB [Construction]
∴By Basic Proportionality Theorem, we have :0
ED/AE = DO/BO
or AE/ED = BO/DO … (2)
From (1) and (2), we obtain that
AO/CO = BO/DO
∴ AO/BO = CO/DO.
Question 10.
The diagonals of a quadrilateral ABCD intersect each other at the point O such that AO/BO = CO/DO that ABCD is a trapezium.
Solution:
Given : A quadrilateral ABCD in which its diagonals AC and BD intersect each other at the point O such that AO/BO = CO/DO, i.e., AO/CO = BO/DO
To prove : Quadrilateral ABCD is a trapezium.
Construction : Through O, draw OE || AB meeting AD in E.
Proof:
In ∆ ADB, we have :
OE || AB [Construction]
∴ By Basic Proportionality Theorem, we have :
DE/EA = OD/BO
or EA/DE = BO/DO
i.e., EA/DE = BO/DO
∴ EA/DE = AO/CO [ ∵AO/CO = BO/DO(given)]
Thus, in ∆ ADC, points E and O are dividing the sides AD and AC in the same ratio. Therefore, by the converse of Basic Proportionality Theorem, we have :
EO || DC
But, EO || AB [construction]
Hence, AB || DC
∴ Quadrilateral ABCD is a trapezium
