Question

ABCD is trapezium in which AB||DC and it's diagonals intersect each other at the point O. Show that AO/BO = CO/DO ​

Answer 1

Given :

ABCD is a trapezium and AB║DC

To Prove :

AO/BO = CO/DO

Construction :

Draw OE║DC such that E lies on BC

Proof :

• In △BDC

→ BO/OD = BE/EC....(1)     [By Basic Proportionality Theorem]

• In △ABC

→ AO/OC = BE/EC.....(2)    [By Basic Proportionality Theorem]

From Equation (1) and (2) :

→ AO/OC = BO/OD

→ $\boxed{AO/BO = CO/OD}$

→ HENCE PROVED ←

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.