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In Fig 6.17, if PQRS is a parallelogram and AB||PS, then prove that OC||SR

In Fig 6.17, if PQRS is a parallelogram and ABPS, then prove that OCSR

Solution:

Given, PQRS is a parallelogram

Also, AB||PS

We have to prove that OC||SR.

In △OPS and △OAB,

∠POS = ∠AOB = common angle

The corresponding angles are equal. i.e.,∠OSP = ∠OBA

AAA criterion states that if two angles of a triangle are respectively equal to two angles of another triangle, then by the angle sum property of a triangle their third angle will also be equal.

By AAA criterion, △OPS ⩬ △OAB

Basic Proportionality Theorem (BPT) states that if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.

By BPT, PS/AB = OS/OB ----------------- (1)

In △CQR and △CAB,

∠RCQ = ∠BCA = common angle

The corresponding angles are equal. i.e.,∠CRQ = ∠CBA

By AAA criterion, △CQR ⩬ △CAB

By BPT, QR/AB = CR/CB ----------------- (2)

Given, PQRS is a parallelogram PS = QR,

So (2) becomes PS/AB = CR/CB ------------ (3)

From (2) and (3),

OS/OB = CR/CB

On rearranging,

OB/OS = CB/CR

On subtracting 1 from both sides, we get,

OB/OS- 1 = CB/CR - 1

OB-OS / OS = CB-CR / CR

From the figure,

OB - OS = BS

CB - CR = BR

Now, BS/OS = BR/CR

By Converse of Basic Proportionality Theorem,

SR||OC

Therefore, it is proven that OC||SR

✦ Try This: If Q is a point oh the side SR of a ΔPSR such that PQ = PR, then prove that PS > PQ.

☛ Also Check: NCERT Solutions for Class 10 Maths Chapter 6

NCERT Exemplar Class 10 Maths Exercise 6.4 Problem 4

In Fig 6.17, if PQRS is a parallelogram and AB||PS, then prove that OC||SR

Summary:

In Fig 6.17, if PQRS is a parallelogram and AB||PS, then it is proven that OC||SR by converse of Basic Proportionality Theorem which states that if a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side

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