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Answer :

Given, in \(\triangle ABC\), D and E are the mid points of AB and AC respectively, such that,

AD=BD and AE=EC.

We have to prove that: DE || BC.

Since, D is the midpoint of AB

AD=DB

=>\(\frac{AD}{BD}\) = 1……………………………….. (i)

Also given, E is the mid-point of AC.

AE=EC

=> \(\frac{AE}{EC}\) = 1

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

\(\frac{AD}{BD} = \frac{AE}{EC}\)

By converse of Basic Proportionality Theorem,

DE || BC

Hence, proved.

- 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).
- 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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