Math, asked by krishnandecorator59, 6 months ago

please answer this question fast ​

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Answered by TanviJ
1

Answer:

here is ur ans

Step-by-step explanation:

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

We have to prove, AO/BO = CO/DO

From the point O, draw a line EO touching AD at E, in such a way that,

EO || DC || AB

In ΔADC, we have OE || DC

Therefore, By using Basic Proportionality Theorem

AE/ED = AO/CO ……………..(i)

Now, In ΔABD, OE || AB

Therefore, By using Basic Proportionality Theorem

DE/EA = DO/BO…………….(ii)

From equation (i) and (ii), we get,

AO/CO = BO/DO

⇒AO/BO = CO/DO

Hence, proved.

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