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

1 Answer

Solution :-

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

Ncert solutions class 10 chapter 6-12

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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