Question

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

Answer 1

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 $\frac{\mathrm{AO}}{\mathrm{BO}}=\frac{\mathrm{CO}}{\mathrm{DO}}$