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Answer :

(i) Given, in Triangle ABC, DE II BC

\({{AD} \over {DB}} = {{AE} \over {EC}}\)

[Using Basic proportionality theorem]

\(\Rightarrow {{1.5} \over {3}} = {{1} \over {EC}}\)

\(\Rightarrow EC = {{3} \over {1.5}}\)

\(\Rightarrow \)EC = 3×\({{10} \over {15}}\) = 2 cm

Hence, EC = 2 cm.

(ii) Given, in Triangle ABC, DE is parallel to BC

= \({{AD} \over {DB}} = {{AE} \over {EC}}\)

[Using Basic proportionality theorem]

\(\Rightarrow {{AD} \over {7.2}}= {{1.8} \over {5.4}}\)

\(\Rightarrow AD = 1.8 ×{{7.2} \over {5.4}} \)

\( = ({{18} \over {10}})×({{72} \over {10}})×({{10} \over {54}}) \)

\( = {{24} \over {10}}\)

AD = 2.4 cm

Hence, AD = 2.4 cm.

- 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 = 0.63 cm
- In the figure, if LM || CB and LN || CD, prove that \(\frac{AM}{AB} = \frac{AN}{AD}\).
- In the figure, DE||AC and DF || AE. Prove that \(\frac{BF}{FE} = \frac{BE}{EC}\).
- In the figure, DE||OQ and DF||OR, show that EF||QR.
- 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.
- Using Basic proportionality theorem, prove that a line drawn through the mid-points of one side of a triangle parallel to another side bisects the third side.(Recall that you have proved it in Class IX).
- Using Converse of basic proportionality theorem, 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).
- ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O. Show that \(\frac{AO}{BO} = \frac{CO}{DO}\).
- The diagonals of a quadrilateral ABCD intersect each other at the point O such that \(\frac{AO}{BO} = \frac{CO}{DO}\). Show that ABCD is a trapezium.

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