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Sia ? 6 years, 6 months ago
Given: In ABC, {tex}\frac { \mathrm { AP } } { \mathrm { BP } } = \frac { \mathrm { AQ } } { \mathrm { QC } }{/tex}
To prove: PQ {tex}\|{/tex} BC
Construction: Let PE {tex}\|{/tex} BC(construction).
{tex}\therefore \quad \frac { \mathrm { AP } } { \mathrm { PB } } = \frac { \mathrm { AE } } { \mathrm { EC } }{/tex} ..(i)
Also, {tex}\frac { \mathrm { AP } } { \mathrm { BP } } = \frac { \mathrm { AQ } } { \mathrm { QC } }{/tex} [given (ii)]
From (i) and (ii) {tex}\frac { \mathrm { AE } } { \mathrm { EC } } = \frac { \mathrm { AQ } } { \mathrm { QC } } \Rightarrow{/tex} E and Q coincides
{tex}\therefore{/tex} PQ {tex}\|{/tex} BC.
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