Question

ABCD is a trapezium in which AB parallel to DC and its diagonals intersect each other at the point O. show that AO÷BO=CO÷DO

Answer 1

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Answer 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,

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

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

Hence Proved.