Question

In the given figure , OA = OB and op =0Q. Prove that (i)px=QX (ii). AX = BX
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Answer 1

Given :

OA = OB

OP = OQ

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To Prove :

PX = QX

AX = BX

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ProoF :

In ∆ OAQ and ∆ OPB , we have

→ OA = OB ---------------------- ( Given )

→ ∠O = ∠O ------------- ( Common Angle )

→ OQ = OP ----------------------- ( Given )

So, ∆ OAQ ≅ ∆ OPB ( SAS congruency )

∠OBP = ∠OAQ ------------------- ( 1 )

Thus, in ∆BXQ and ∆PXA, we have,

BQ = OB – OQ

And PA = OA – OP

But OP = OQ

And OA = OB …given

Hence, we have, BQ = PA --------------- ( 2 )

Now consider ∆BXQ and ∆PXA,

∠BXQ = ∠PXA ------ (vertically opposite angles)

∠OBP = ∠OAQ ------------------ ( From 1 )

BQ = PA ----------------------------- ( From 2 )

Thus by AAS property of congruence,

∆BXQ ≅ ∆PXA

Hence, we know that, corresponding parts of the congruent triangles are equal

∴ PX = QX

And AX = BX