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Q. 74.4( 9 Votes )

# In the given figu

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))

2QOS =180°

QOS=90°

Hence, SO is the perpendicular bisector of QR

So, SP is also the perpendicular bisector of QR

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