The diagonals of a trapezium divide each other proportionally. Prove i

1.	The diagonals of a trapezium divide each other proportionally. Prove it
Answer» Given in the quadrilateral ABCD{tex}\\frac{AO}{BO}=\\frac{CO}{DO}{/tex}or,\xa0{tex}\\frac { A O } { C O } = \\frac { B O } { D O }{/tex} ...(i)Draw\xa0EO {tex}\\parallel{/tex} AB onIn\xa0{tex}\\triangle A B D,{/tex}\xa0EO {tex}\\parallel{/tex} AB (By construction){tex}\\therefore {/tex}\xa0{tex}\\frac { A E } { E D } = \\frac { B O } { D O }{/tex}\xa0(By BPT)...(ii)From (i) and (ii) we get\xa0{tex}\\frac{AO}{CO}=\\frac{AE}{ED}{/tex}Hence by converse of BPT in {tex}\\triangle{/tex}ADC\xa0{tex}EO\\\|CD{/tex}But\xa0{tex}EO \\\|AB {/tex}So {tex}AB\\\|CD {/tex}Therefore ABCD is a trapezium

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