# PQRS is a trapezium such that PQ||SR. The ratio PR:QS is always equal to 1) PQ:SR 2) PO-OR:QO-OS 3) PO+OR:QO-OS 4) PO^2:OQ^2

Solution

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$\mathrm{In}△\mathrm{POQ}\mathrm{and}△\mathrm{ROS}\phantom{\rule{0ex}{0ex}}\angle \mathrm{POQ}=\angle \mathrm{ROS}\left[\mathrm{vertically}\mathrm{opposite}\mathrm{angle}\right]\phantom{\rule{0ex}{0ex}}\angle \mathrm{OPQ}=\angle \mathrm{ORS}\left[\mathrm{alternate}\mathrm{interior}\mathrm{angle}\right]\phantom{\rule{0ex}{0ex}}\therefore △\mathrm{POQ}\cong △\mathrm{ROS}\left[\mathrm{By}\mathrm{AA}\mathrm{criterion}\right]\phantom{\rule{0ex}{0ex}}\mathrm{Hence},\frac{\mathrm{PO}}{\mathrm{RO}}=\frac{\mathrm{OQ}}{\mathrm{OS}}=\frac{\mathrm{PQ}}{\mathrm{RS}}\left[\mathrm{CPCT}\right]...\left(1\right)\phantom{\rule{0ex}{0ex}}⇒\frac{\mathrm{PO}}{\mathrm{RO}}=\frac{\mathrm{OQ}}{\mathrm{OS}}\phantom{\rule{0ex}{0ex}}⇒\frac{\mathrm{PO}}{\mathrm{RO}}+1=\frac{\mathrm{OQ}}{\mathrm{OS}}+1\phantom{\rule{0ex}{0ex}}⇒\frac{\mathrm{PO}+\mathrm{RO}}{\mathrm{RO}}=\frac{\mathrm{OQ}+\mathrm{OS}}{\mathrm{OS}}\phantom{\rule{0ex}{0ex}}⇒\frac{\mathrm{PR}}{\mathrm{RO}}=\frac{\mathrm{QS}}{\mathrm{OS}}\phantom{\rule{0ex}{0ex}}⇒\frac{\mathrm{PR}}{\mathrm{QS}}=\frac{\mathrm{RO}}{\mathrm{OS}}\phantom{\rule{0ex}{0ex}}⇒\frac{\mathrm{PR}}{\mathrm{QS}}=\frac{\mathrm{PQ}}{\mathrm{RS}}\left[\mathrm{using}1\right]$
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