Math, asked by kvbegumpet13, 3 months ago

help me in tjis question​

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Answered by prakharjalonha95
0

hope it's helpful to you

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Answered by DeeznutzUwU
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       \underline{\bold{Given:}}

       TS \parallel QR

       TV \parallel QS

       \underline{\bold{To-Prove:}}

       \dfrac{PV}{VS}= \dfrac{PS}{SR}

       \underline{\bold{Proof:}}

       \text{In }\triangle{PQR} \text{ and }\triangle{PTS}

\implies \angle{PTS}= \angle{PQR}\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }(TS\parallel QR)\text{(Corresponding Angles)}

\implies \angle{PST}= \angle{PRQ}\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }(TS\parallel QR)\text{(Corresponding Angles)}

 \therefore \text{ }\text{ By AA similarity }\triangle {PTS} \text{ is similar to }\triangle {PQR}      

\implies \dfrac{PS}{PR} = \dfrac{PT}{PQ} \text{ ------ (i)} \text{ }\text{ }\text{ }\text{ }\text{ }\text{ }(\triangle {PTS} \text{ is similar to }\triangle {PQR})

       \text{In }\triangle{PTV} \text{ and }\triangle{PQS}

\implies \angle{PTV}= \angle{PQS}\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }(TV\parallel QS))\text{(Corresponding Angles)}

\implies \angle{PVT}= \angle{PSQ}\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }(TV\parallel QS))\text{(Corresponding Angles)}

 \therefore \text{ }\text{ By AA similarity }\triangle {PTV} \text{ is similar to }\triangle {PQS}      

\implies \dfrac{PV}{PS} = \dfrac{PT}{PQ} \text{ ------ (ii)} \text{ }\text{ }\text{ }\text{ }\text{ }\text{ }(\triangle {PTV} \text{ is similar to }\triangle {PQS})

       \text{From (i) and (ii)}

\implies \dfrac{PS}{PR} = \dfrac{PV}{PS}

       \text{Taking reciprocal on both sides}

\implies \dfrac{PR}{PS} = \dfrac{PS}{PV}

       \text{Substracting 1 on both sides}

\implies \dfrac{PR}{PS} -1= \dfrac{PS}{PV} -1

       \text{Simplifying...}

\implies \dfrac{PR-PS}{PS} = \dfrac{PS-PV}{PV}

       \text{We know that, }(PR-PS) = SR\text{ and }(PS-PV) = VS

\implies \dfrac{SR}{PS} = \dfrac{VS}{PV}

       \text{Taking reciprocal on both sides}

\implies \dfrac{PS}{SR} = \dfrac{PV}{VS}

       \text{Hence Proved}

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