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Answer :

Given, Quadrilateral ABCD where AC and BD intersects each other at O such that,

\(\frac{AO}{BO} = \frac{CO}{DO}\).

We have to prove here, ABCD is a trapezium

From the point O, draw a line EO touching AD at E, in such a way that,

EO || DC || AB

In \(\triangle DAB\), EO || AB

Therefore, By using Basic Proportionality Theorem

\(\frac{DE}{EA} = \frac{DO}{OB}\) ……………………(i)

Also, given,

\(\frac{AO}{BO} = \frac{CO}{DO}\)

=> \(\frac{AO}{CO} = \frac{BO}{DO}\)

=> \(\frac{CO}{AO} = \frac{DO}{BO}\)

=> \(\frac{DO}{OB} = \frac{CO}{AO}\) …………………………..(ii)

From equation (i) and (ii), we get

\(\frac{DE}{EA} = \frac{CO}{AO}\)

Therefore, By using converse of Basic Proportionality Theorem,

EO || DC also EO || AB

=> AB || DC.

Hence, quadrilateral ABCD is a trapezium with AB || CD.

- In figure. (i) and (ii), DE || BC. Find EC in (i) and AD in (ii).
- 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 \(\frac{AO}{BO} = \frac{CO}{DO}\).

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