Math, asked by prasoon660, 1 year ago

ABCD is a trapezium in which ab parallel DC and its diagonals intersect each other at a point O show that AO/BO=CO/DO

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Answered by Anonymous
14
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Answered by BlessedMess
2

Given,

  • ABCD is a trapezium where AB||DC and diagonals AC and BD intersect each other at O.

To prove,

  • \large\sf{\dfrac{AO}{BO}=\dfrac{CO}{DO}}

From the point O,draw a line EO touching AD at E,in such a way that,EO||DC||AB

In triangle ADC,we have OE||DC

Therefore, by using basic proportionality theorem

\large\sf{\frac{AE}{ED}  =  \frac{AO}{CO}}..............(i)

Now,in triangle ABD OE||AB

By using basic proportionality theorem

\large\sf{\frac{DE}{EA}  =  \frac{DO}{BO}}..............(ii)

From equation (i) and (ii), we get,

\large\sf{\frac{AO}{CO}  =  \frac{BO}{DO}}

\large\sf{→\frac{AO}{BO}  =  \frac{CO}{DO}}

Hence Proved.

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