In the given figure two isosceles triangle PQR and SQR lie on the same side of the common base QR. Prove that line SP is the perpendicular bisector of line QR.

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