In trapezium $A B C D$ , side $A B ∥ C D$ and diagonals $A C$ and $B D$ intersect each other at $O$ .
Prove that: $\frac{O A}{O C} = \frac{O D}{O B}$ .

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Solution

In trapezium $A B C D$ , side $A B ∥ C D$

Diagonals $A C$ and $B D$ intersect each other at point $O$ .

To prove: $\frac{O A}{O C} = \frac{O D}{O B}$

Proof:

In $△ A O B$ and $△ C O D$

$∠ O A B = ∠ O C D$ ..........(Alternate angles)

and $∠ A O B = ∠ C O D$ ........(Vertically opposite angles)

Hence, $△ A O B \sim$ $△ C O D$ .......(By AA test of similarity)

$\Rightarrow$ $\frac{O A}{O C} = \frac{O D}{O B}$ ........[C.S.S.T] $[h e n c e p r o v e d]$

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