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प्रश्न
In parallelogram PQRS, A and B are points on PR such that PA = AB = BR. Prove that ASBQ is a parallelogram.

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उत्तर
Given:
- PQRS is a parallelogram.
- A and B are points on diagonal PR such that PA = AB = BR.
To Prove: ASBQ is a parallelogram.
Proof / Derivation:
1. In parallelogram PQRS, the diagonal PR is divided into three equal segments by points A and B, i.e., PA = AB = BR.
2. Since PQRS is a parallelogram, opposite sides are parallel and equal:
PS || QR and PS = QR
PQ || SR and PQ = SR.
3. Because A and B divide PR into three equal parts, vectors can express the points as: Let vector representations be:
`P = vecP `
`R = vecR`
Then A divides PR so that `vecA = vecP + 1/3 (vecR - vecP)`
B divides PR so that `vecB = vecP + 2/3 (vecR - vecP)`
4. We want to prove ASBQ is a parallelogram. For that, we can prove vectors `vec(AS)` and `vec(BQ)` are equal. Then AS || BQ and AS = BQ (one pair of opposite sides of a parallelogram).
5. Find `vec(AS) : vec(AS) = vecS - vecA`
6. Find `vec(BQ) : vec(BQ) = vecQ - vecB`
7. Using the property of the parallelogram, `vecS = vecP + vecQ - vecR` (since PQRS is a parallelogram)
8. Substitute these in:
`vec(AS) = vecS - vecA`
= `(vecP + vecQ - vecR) - (vecP + 1/3 (vecR - vecP))`
= `vecQ + vecP - vecR - vecP - 1/3 vecR + 1/3 vecP`
= `vecQ - 2/3 vecR + 1/3 vecP`
`vec(BQ) = vecQ - vecB`
= `vecQ - (vecP + 2/3 (vecR - vecP))`
= `vecQ - vecP - 2/3 vecR + 2/3 vecP`
= `vecQ - 2/3 vecR + (-1 + 2/3)/vecP`
= `vecQ - 2/3 vecR + 1/3 vecP`
(Note: The algebra simplifies to the same vector for `vec(AS)` and `vec(BQ)`.
9. Since `(vec(AS) = vec(BQ))`, this implies AS || BQ and AS = BQ.
10. In quadrilateral ASBQ, opposite sides AS and BQ are equal and parallel.
11. Similarly, since AB and SQ are parts of the diagonal and side respectively, and PQRS is a parallelogram, it follows by the properties that AB || SQ and AB = SQ.
12. Therefore, both pairs of opposite sides of ASBQ are parallel and equal.
ASBQ is a parallelogram because both pairs of its opposite sides are equal and parallel.
