Question

State and prove Basic Proportionality Theorem. Using the above theorem, if ABCD is a trapezium whose diagonals intersect each other at O show that AO/OC = BO/OD.​

Answer 1

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Given parameters

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

To prove

[tex] \frac{AO}{BO} [/tex] = [tex] \frac{CO}{DO} [/tex]

 

 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

[tex] \frac{AE}{ED} [/tex] = [tex] \frac{AO}{OC} [/tex]………………………………(1)

Now consider Δ ABD

where EO ll AB

According to basic proportionality theorem

[tex] \frac{AE}{ED} [/tex] = [tex] \frac{BO}{OD} [/tex]……………………………..(2)

From equation (1) and (2) we have

[tex] \frac{AO}{OC} [/tex] = [tex] \frac{BO}{OD} [/tex]

 

 Hence the proof.

Answer 2

Given parameters

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

To prove

AO/BO=CO/DO

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

Given parameters

AE/ED=AO/OC………………………………(1)

Now consider Δ ABD

where EO ll AB

According to basic proportionality theorem

AE/ED=BO/OD……………………………..(2)

From equation (1) and (2) we have

AO/OC=BO/OD

Hence the proof.