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In Fig. 6.17, (i) and (ii), DE || BC. Find EC in (i) and AD in (ii).
(i) 
Let EC = x cm
It is given that DE || BC.
By using basic proportionality theorem, we obtain
(ii) Let AD = x cm
It is given that DE || BC.
By using basic proportionality theorem, we obtain
In Fig., DE || OQ and DF || OR. Show that EF || QR.
In Δ POQ, DE || OQ
(basic proportionality theorem)
From (i) and (ii) we get,
(converse of BPT)
In Fig., A, B and C are points on OP, OQ and OR respectively such that AB || PQ and AC || PR. Show that BC || QR.
In Δ POQ, AB || PQ
From (i) and (ii) we get
E and F are points on the sides PQ and PR respectively of a triangle 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
(i) 
Given that, PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm, FR = 2.4 cm
Therefore, EF is not parallel to QR
(ii) 
PE = 4 cm, QE = 4.5 cm, PF = 8 cm, RF = 9 cm
therefore, EF is parallel to QR.
PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm, PF = 0.36 cm
Hence
Therefore, EF is parallel to QR.
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).
Consider the given figure in which PQ is a line segment joining the mid-points P and Q of line AB and AC respectively.
i.e., AP = PB and AQ = QC
It can be observed that and
Hence, by using basic proportionality theorem, we obtain
ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O. Show that
Draw a line EF through point O, such that
In ΔADC,
By using basic proportionality theorem, we obtain
In
So, by using basic proportionality theorem, we obtain
The diagonals of a quadrilateral ABCD intersect each other at the point O such that ⋅ Show that ABCD is a trapezium.
Consider the following figure
Draw a line OE || AB
In ΔABD,
By using basic proportionality theorem, we obtain
⇒ EO || DC [By the converse of basic proportionality theorem]
⇒ AB || OE || DC
⇒ AB || CD
∴ ABCD is a trapezium.
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