Assertion (A): The straight line drawn through the mid-point of one side of a triangle and parallel to another side bisects the third side.

Reason (R): If a transversal makes equal intercepts on three or more parallel lines, then any other line cutting them will also make equal intercepts.

1. A is true, R is false.
2. A is false, R is true.
3. Both A and R are true.
4. Both A and R are false.

Both A and R are true.

Explanation

Given: In triangle ABC, D is mid-point of side AB and DE is parallel to BC.

To Prove: DE bisects AC, i.e. AE = EC

Construction: Through vertex A, draw FG parallel to BC so that FG ∥ BC ∥ DE.

Proof:

Since, FG || BC || DE and the traversal AB makes equal intercepts on these three parallel lines i.e. AD = DB.

Also, AC is an another traversal. According to Intercept Theorem, if a traversal makes equal intercepts on three or more parallel lines, then any other traversal, for the same parallel lines, will also make equal intercepts.

∴ AE = CE

∴ Assertion (A) is true.

Given : Traversal AB makes equal intercepts on three parallel lines l, m and n.

i.e., l || m || n and PQ = QR

CD is another traversal which makes intercepts LM and MN.

To Prove : LM = MN

Construction : Draw PS and QT parallel to CD.

Proof : In △ PQS and △ QRT,

PQ = QR (Given)

∠ PQS = ∠ QRT (Corresponding angles)

∠ QPS = ∠ RQT (Corresponding angles as PS || CD || QT)

∴ △ PQS ≅ △ QRT (A.S.A.)

∴ PS = QT (C.P.C.T.)

As both the pairs of opposite sides are parallel. So, PSML is a parallelogram.

∴ PS = LM (opposite sides of parallelogram are equal.)

As both the pairs of opposite sides are parallel. So, QTNM is a parallelogram.

∴ QT = MN (opposite sides of parallelogram are equal.)

∴ LM = MN

∴ Reason (R) is true.

Hence, both Assertion (A) and Reason (R) are true.