Question

help me in tjis question​

Answer 1

hope it's helpful to you

Answer 2

       $\underline{\bold{Given:}}$

       $TS \parallel QR$

       $TV \parallel QS$

       $\underline{\bold{To-Prove:}}$

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

       $\underline{\bold{Proof:}}$

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

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

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

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

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

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

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

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

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

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

       $\text{From (i) and (ii)}$

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

       $\text{Taking reciprocal on both sides}$

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

       $\text{Substracting 1 on both sides}$

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

       $\text{Simplifying...}$

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

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

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

       $\text{Taking reciprocal on both sides}$

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

       $\text{Hence Proved}$