Answer :

Given that ∆PQR and ∆SQR lie on the same side of the common base QR

Also, ∆PQR and ∆SQR are isosceles triangles .

So, QS=RS and QP=RP…………(1)

To prove : SP is the perpendicular bisector of line QR

Proof : In ∆PQS and ∆PRS,

QS=RS (given)

QP=RP (given)

SP=SP (common)

∴ ∆PQS ∆PRS (by SSS rule)

So, ∠PSQ=∠PSR (by cpct)…………….(2)

Now, in ∆QSO and ∆RSO,

QS=RS (given)

SO=SO (common)

∠PSQ=∠PSR (from (2))

∴ ∆QSO ∆RSO (by SAS rule)

So, QO=RO (by cpct)

And ∠QOS=∠ROS (by cpct)………..(3)

Now, ∠QOS+∠ROS=180° (straight angle)

⇒ ∠QOS+∠QOS=180° (from (3))

⇒ 2∠QOS =180°

⇒ ∠QOS=90°

Hence, SO is the perpendicular bisector of QR

So, SP is also the perpendicular bisector of QR

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In the given figuRajasthan Board Mathematics

In the given figuRajasthan Board Mathematics

Prove that the trRajasthan Board Mathematics