Question

ABCD is a trapezium in which AB is Parallel to DC and it's diagonal intersect each other at 'O' show that AO/BO=CO/DO​

Answer 1

Step-by-step explanation:

Answer in attachment pls.

Answer 2

Question :

ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O.Show AO/BO = CO/DO

Solution :

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.