Question

Pls prove fast someone help​

Answer 1

Step-by-step explanation:

hey here is your proof

pls mark it as brainliest

$so \: here \: ts \: parallel \: to \: qr \: and \: tv \: parallel \: to \: qs$

$so \: thus \: as \: ts \: parallel \: to \: qr \\ by \: basic \: proportionality \: theorem \\ we\: get \\ pt \div tq = ps \div sr \: \: \: \: \: (1)$

$also \: considering \: triangle \: ptv \: and \: triangle \: pqs \\ angle \: tpv = angle \: qps \: \: \: \: (common \: angle) \\ since \: tv \: parallel \: to \: qs \\ angle \: ptv = angle \: pqs \: \: \: \: (corresponding \: angles \: theorem) \\ \\ thus \: triangle \: ptv \: similar \: to \: triangle \: pqs \: \: \: \: \: (a.a \: test \: of \: similarity)$

$so \: hence \\ pt \div tq = pv \div vs \: \: \: \: \: \: \: (2)$

$thus \: from \: (1) \: and \: (2) \\ we \: get \\ pv \div vs = ps \div sr \\ hence \: proved$

Answer 2

hope it is helpful to you

by zubi✌️✌️✌️

yeah it is me only

how r u?