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 parameters

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

To prove

AOBO=CODO

Construction

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

Now, AB ll CD

By construction EF ll AB

∴ EF ll CD

Consider the ΔADC,

Where EO ll AB

According to basic proportionality theorem

AEED=AOOC ………………………………(1)

Now consider Δ ABD

where EO ll AB

According to basic proportionality theorem

AEED=BOOD ……………………………..(2)

From equation (1) and (2) we have

AOOC=BOOD

⇒ AOBO=OCOD

Hence the proof.

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Answer 2

Correct 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.