Question

Hola Mates ! ❤️

Question :
ABCD is a trapezium in which ABIIDC and it's diagonals intersect each other at point 'O'.
Show that AO/BO = CO/DO

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Answer 1

$\huge\mathfrak{Answer:}$

Given:

• We have been given a trapezium ABCD such that AB || DC.
• The diagonals AC and BD intersect each other at O.

Construction:

• Let us draw OE parallel to AB or DC.

Solution:

We have been given a trapezium ABCD such that AB || DC. The diagonals AC and BD intersect each other at O.

We have constructed OE || DC.

Now,

OE || DC [By construction]

∴ Using the Basic Proportionality Theorem, we have

$\sf{ \dfrac{AE}{ED} = \dfrac{AO}{CO} }$______________(1)

Now in △ABD,

OE || AB [ By Construction]

∴ Using the Basic Proportionality Theorem, we have

$\sf{ \dfrac{ED}{AE} = \dfrac{DO}{BO} }$

$\implies\sf{ \dfrac{AE}{ED} = \dfrac{BO}{DO} }$________(2)

Now, from equation (1) and (2) we get,

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

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

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

Answer 2

$\huge\mathfrak\blue{Answer:}$

Given:

• We have been given a Trapezium ABCD in which AB II CD
• The Diagonals AC and BD intersect at point O

To prove :

• $\dfrac{AO}{BO}=\dfrac{CO}{DO}$

Construction:

• Let us draw a line segment OX Parallel to AB or CD
• OX II AB or OX II CD ---------( 1 )

Solution:

Since it is given that the Diagonals AC and BD of Trapezium ABCD intersect at point O.

Now , In ∆ ABC

OX II AB [ Equation 1 ]

Using Thales Theorem , we get

$\dfrac{AO}{CO}=\dfrac{BX}{XC}$ ------------------ ( 2 )

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Now In ∆ BCD

OX II CD [ Equation 1 ]

Using Thales Theorm , we get

$\dfrac{BX}{XC}=\dfrac{BO}{DO}$ ----------------- ( 3 )

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From Equation ( 2 ) and ( 3 ) we get

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

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

Hence Proved !!