Question

9. ABCD is a trapezium in which AB || DC and its
diagonals intersect each other at the point O. Show
AO _co
that BO
DO​

Answer 1

Answer:Given: □ABCD is a trapezium where, AB ll CD

Diagonals AC and BD intersect at point O.

Construction: Draw a line EF passing through O and also parallel to AB.

Now, AB ll CD, since by construction, EF ll AB ⇒ EF ll CD

Consider the ΔADC,

EO ll DC

Thus, by Basic proportionality theorem, (AE / ED) = (AO / OC) .... (i)

Now, consider Δ ABD,

EO ll AB,

Thus, by Basic proportionality theorem, (AE / ED) = (BO / OD) .... (ii)

From (i) and (ii), we have, (AO / OC) = (BO / OD) (since L.H.S of i and ii are equal)

Hence we proved that, (AO / OC) = (BO / OD)

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

Given :

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

To prove :

$\sf{\dfrac{AO}{BO}=\dfrac{CO}{DO}}$

Solution :

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

In triangle ADC,we have OX||DC

Therefore, by using basic proportionality theorem

$\sf{\frac{AE}{XD} = \frac{AO}{CO}}$--(i)

Now,in triangle ABD OX||AB

By using basic proportionality theorem

$\sf{\frac{DX}{XA} = \frac{DO}{BO}}$--(ii)

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

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

$\sf{→\frac{AO}{BO} = \frac{CO}{DO}}$

Hence Proved.

Additional Information :

Basic proportionality theorem :

If a line is drawn parallel to one side of the triangle , Then the other sides are divided in the same ratio.

Here, We prove that

In trapezium ABCD , AO/BO = CO/DO

Using the Basic proportionality theorem.

We constructed OX || AB and proceeded with the problem.

Check out the attachment for detailed explanation.