Math, asked by Stanza5591, 1 year ago

Abcd is a trapezium in which ab parallel to dc and it's diagonals intersects each other at the pot o show that ao/bo =co/do

Answers

Answered by soteyashaswini
0
Hoping my answer is correct
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Answered by Anonymous
57

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.

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