Given, In ΔPQO, DE || OQ
By using Basic Proportionality Theorem,
$\dfrac{PD}{DO} = \dfrac{PE}{EQ}$……………… ..(i)
Again given, in ΔPOR, DF || OR,
Using Basic Proportionality Theorem,
$\dfrac{PD}{DO} = \dfrac{PF}{FR}$………………… (ii)
From equation (i) and (ii), we get,
$\dfrac{PE}{EQ} = \dfrac{PF}{FR}$
Therefore, by converse of Basic Proportionality Theorem,
EF || QR, in ΔPQR.
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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.
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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). |
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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). |
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ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O. Show that $\dfrac{AO}{BO} = \dfrac{CO}{DO}$ |
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The diagonals of a quadrilateral ABCD intersect each other at the point O such that $\dfrac{AO}{BO} = \dfrac{CO}{DO}$ Show that ABCD is a trapezium. |
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In figure. (i) and (ii), DE || BC. Find EC in (i) and AD in (ii).
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E and F are points on the sides PQ and PR respectively of a Δ PQR. (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 |
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In Fig. 6.18, if LM || CB and LN || CD, prove that $\dfrac{AM}{AB} = \dfrac{ AN}{AD}$
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In Fig. 6.19, DE || AC and DF || AE. Prove that $\dfrac{BF}{FE} = {BE}{EC}$ |
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In Fig. 6.20, DE || OQ and DF || OR. Show that EF || QR.
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