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# $\mathrm{ABCD}$ is a trapezium in which $\mathrm{AB}$ parallel $\mathrm{DC}$and its diagonals intersect each other at point $\mathrm{O}$ . show that $\frac{\mathrm{AO}}{\mathrm{BO}}=\frac{\mathrm{CO}}{\mathrm{DO}}$

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## Step-1 Construction: Draw a line $\mathrm{EF}$ passing through $\mathrm{O}$ and also parallel to $\mathrm{AB}$Now, $\mathrm{AB}\mathrm{ll}\mathrm{CD}$$\mathrm{EF}\mathrm{ll}\mathrm{AB}$ by Construction$\therefore \mathrm{EF}\mathrm{ll}\mathrm{CD}$Step-2 Basic Proportionality theorem:Consider the $\mathrm{\Delta ADC}$Where $\mathrm{EO}\mathrm{ll}\mathrm{AB}$According to the basic proportionality theorem$\frac{\mathrm{AE}}{\mathrm{ED}}=\frac{\mathrm{AO}}{\mathrm{OC}}$ …………………(i)Now consider $\mathrm{\Delta }\mathrm{ABD}$where $\mathrm{EO}\mathrm{ll}\mathrm{AB}$According to the basic proportionality theorem$\frac{\mathrm{DE}}{\mathrm{EA}}=\frac{\mathrm{DO}}{\mathrm{BO}}$…………….(ii)From equation (i) and (ii), we get,$\frac{\mathrm{AO}}{OC}=\frac{\mathrm{BO}}{\mathrm{DO}}$⇒$\frac{\mathrm{AO}}{\mathrm{BO}}=\frac{\mathrm{CO}}{\mathrm{DO}}$Hence, proved.Therefore we proved that $\frac{\mathrm{AO}}{\mathrm{BO}}=\frac{\mathrm{CO}}{\mathrm{DO}}$  Suggest Corrections  23      Similar questions  Explore more