* General enunciation:* We have to prove that if the opposite sides of a quadrilateral are equal and parallel, it is a

*parallelogram.*

* Particular enunciation*: Let the opposite sides of the quadrilateral ABCD are equal and parallel i.e., BC = AD, BC || AD and AB = CD, AB || CD.

We have to prove that ABCD is a *parallelogram.*

Construction: We join A, C and B, D.

Proof: In **∆**ABC and **∆**ADC.

AB = DC, BC = AD [by supposition]

and AC = AC ——————(i) [common side]

∴**∆**ABC ≅ **∆**ADC.

∴ ∠B = ∠D ——————(ii)

In **∆**BAD and **∆**BCD

AB = DC, BC = AD [by supposition]

and BD = BD ——————–(iii) [common side]

∴**∆**BAD ≅ **∆**BCD.

∴ ∠A = ∠C ——————————–(iv)

In the quadrilateral ABCD,

BC = AD, AB = DC [ from (i) and (iii)]

and ∠B = ∠D, ∠A = ∠C [from (ii) and (iv)]

*We know that, the opposite sides and angles of a parallelogram are equal.*

∴ ABCD is a parallelogram. ** (Proved).**

*Problem: The median BO of ***∆**ABC is produced up to D so that BO = OD. Prove that ABCD is a parallelogram.

**∆**ABC is produced up to D so that BO = OD. Prove that ABCD is a parallelogram.

**Solution:** General Enunciation: Problem: The median BO of **∆**ABC is produced up to D so that BO = OD. Prove that ABCD is *parallelogram*.

**Particular enunciation**: Given that the median BO of **∆**ABC is produced up to D so that BO = OD. We join C, D and A, D. We have to prove that ABCD is parallelogram.

Proof: In **∆**ABC, the point of the side AC is O and given that BO is the median of **∆**ABC.

∴ CO = OA ————————(i)

The diagonal of the quadrilateral ABCD is AC and BD where

CO = OA, [ from (i)]

BO = OD. [According to construction]

## i.e., the diagonals bisect each other.

Therefore, ABCD is a parallelogram. ** (Proved)**