Question

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

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Answer 1

Answer:

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

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

To Prove : OC || SR

Proof: 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 .

Answer 2

Answer:

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

To Prove : OC || SR

Proof: 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)

oa/op = ob/os=ab/ps(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}

ca/CQ=cb/cr = ab/qr

PQRS is a parallelogram .

∴PS=QR

\begin{gathered} \frac{ps}{ab} = \frac{cr}{cb} = \frac{cq}{ca} (v) \\ \frac{cr}{cb} = \frac{os}{ob} [From (ii) and (v)] \end{gathered}

ab/ps= cb/cr= ca/cq(v)

cb/cr =ob/os [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 .