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Given : In figure, PR > PQ and PS bisects <QPR.
To Prove : <PSR > <PSQ
Proof : In ∆ PQR
PR > PQ ( Given )
therefore, <PQR > <PRQ.... ( 1 ) (Angle opposite to longer side is greater )
and,. PS is the bisecter of <QPR
and, <QPS = <RPS..... ( 2 )
In ∆ PQS,
<PQS+<QPS+<PSQ = 180° .....( 3 ) ( The sum of three angles of a triangles is 180° )
In ∆ PRS,
<PRS+<SPR+<PSR = 180°.... ( 4 )
Sum of three angles of triangles is 180°. from (3) and (4).
<PQR+<QPS+<PSQ = <PRS+<SPR+<PSR
<PQR+<PSQ = <PQR+<PSR. ( from2 )
<PRS+<PSR = <PQR+<PSQ
<PRS+<PSR = <PRQ+<PSQ. ( from 1 )
<PRQ+<PSR = <PRS+<PSQ. ( <PRQ = <PSR )
<PSR > <PSQ
To Prove : <PSR > <PSQ
Proof : In ∆ PQR
PR > PQ ( Given )
therefore, <PQR > <PRQ.... ( 1 ) (Angle opposite to longer side is greater )
and,. PS is the bisecter of <QPR
and, <QPS = <RPS..... ( 2 )
In ∆ PQS,
<PQS+<QPS+<PSQ = 180° .....( 3 ) ( The sum of three angles of a triangles is 180° )
In ∆ PRS,
<PRS+<SPR+<PSR = 180°.... ( 4 )
Sum of three angles of triangles is 180°. from (3) and (4).
<PQR+<QPS+<PSQ = <PRS+<SPR+<PSR
<PQR+<PSQ = <PQR+<PSR. ( from2 )
<PRS+<PSR = <PQR+<PSQ
<PRS+<PSR = <PRQ+<PSQ. ( from 1 )
<PRQ+<PSR = <PRS+<PSQ. ( <PRQ = <PSR )
<PSR > <PSQ
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