Question

ABCD is a trapezium in which AB ‖ DC and its diagonals intersect each other at the point O. Show that AOBO =CODO​

Answer 1

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=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.

Hope I help you❤❤❤....

Answer 2

Answer:

${ \boxed{ \underline{ \blue{ \sf \: OC / OD = AD / BO }}}}$

Step-by-step explanation:

★ Given :

• ABCD is a trapezium in which AB ‖ DC

★ To Prove :

• AO / BO = CO / DO

★ Solution :

Construction : Draw OM || BA || CD

Proof = In ∆ ABD , AB || MD

: $\implies$ AM / MD = BO / OD ( i )

: $\implies$ Now in ∆ ACD

: $\implies$ OM || CD

: $\implies$ AM / MD = AO / OC ( ii )

: $\implies$ From Eqn ( i ) & ( ii ) Angle

: $\implies$ BO / OD = AO / OC

.: $\implies$ OC / OD = AD / BO

★ Final Answer :

• OC / OD = AD / BO

★ Additional information:

• Before solving this problem the Basic Proportionality Theorem to solve such types of questions. Construction becomes important in solving such questions in a simple manner .

• Draw a line parallel to AB and DC . Using the Basic Proportionality Theorem and the constructed triangles inside the trapezium prove the required answer

• If a line is drawn parallel to one side of a triangle to intersect its other two sides in distinct points, then the other two sides are divided in the same ratio.