ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O. Show that $$\dfrac{ AO}{BO}=\dfrac{CO}{DO}$$.

Asked by Pragya Singh | 1 year ago |  43

##### Solution :-

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

We have to prove, $$\dfrac{AO}{BO}=\dfrac{CO}{DO}$$

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

$$\dfrac{AE}{ED}=\dfrac{AO}{CO}$$ ……………..(i)

Now, In ΔABD, OE || AB

Therefore, By using Basic Proportionality Theorem

$$\dfrac{DE}{EA}=\dfrac{DO}{BO}$$…………….(ii)

From equation (i) and (ii), we get,

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

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

Hence, proved.

Answered by Abhisek | 1 year ago

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