Question

In the figure, PQRS is a //gm and AB //PQ then prove that OC//SR.

Answer this along with step by step explanation
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Answer 1

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This can be proved by applying similarity of triangles and converse of Thales Theorem .

Step-by-step explanation:

To prove - OC║SR

Proof  -  In ΔOPS and ΔOAB

∠POS = ∠AOB     (common in both)

∠OSP = ∠OBA  (corresponding angles are equal as PS║AB)

=> ΔOPS ~ ΔOAB    [AA criteria]

=>  PS/AB =  OS/OB       ........................(1)     (sides in similar triangles are proportional)

In ΔCAB and ΔCRQ

As, QR║AB

=> ∠QCR = ∠ACB        (common)

=>  ∠CBA  = ∠CRQ       (corresponding angles are equal)

=>  ΔCAB ~ ΔCQR           [AA criteria]

=>  CR/CB = QR/AB         (sides in similar triangles are proportional)

Also,  PS = QR        [ PQRS  is parallelogram]

=>  CR/CB = PS/AB                     ......................(2)

From    (1) and (2)

=>   OS/OB =  CR/CB

=>   OB/OS =  CB/CR

Subtracting 1 from both sides

So,  OB/OS - 1 =  CB/CR - 1

=>  (OB - OS)/OS  =  (CB - CR)/CR

=>  BS/OS = BR/CR

By converse of Thales Theorem

=>  OC║SR        .     Hence proved  .

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Answer 2

Answer:

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In the figure, PQRS is a parallelogram and AB||PQ then prove that OC||SR.

$\huge\underline\mathfrak\color{lime}{Solution}$

Given

In ΔABC,PQRS is a parallelogram and PS || AB .

$\large\bf\color{green}{To~prove:}$ OC || SR

In ΔOAB and ΔOPS

PS || AB [Given] (i)

∴ ∠1=∠2

∠3=∠4

∴ΔOPS∼ΔOAB [By AA similarity criterion]

$⇒ \frac{OP}{OA} = \frac{OS}{OB} = \frac{PS}{AB}  (ii)$

PQRS is a parallelogram so PS || QR (iii)

⇒ QR || AB (iv) [From (i) , (iii)]

In ΔCQR and ΔCAB ,

QR || AB [From(iv)]

∠5=∠CAB

∠6=∠CBA [Corresponding angles]

∴ΔCQR∼ΔCAB [By AA similarity criterion]

$⇒ \frac{CQ}{CA} = \frac{CR}{CB} = \frac{QR}{AB}$

PQRS is a parallelogram .

∴PS=QR

$∴ \frac{PS}{AB} = \frac{CR}{CB} = \frac{CQ}{CA}  (v)  \\ ⇒ \frac{CR}{CB} = \frac{OS}{OB}  [From (ii) and (v)]$

These are the ratios of two sides of ΔBOC and are equal so by converse of BPT , SR||OC.

Hence , proved .