Diagonals $A C$ and $B D$ of a trapezium $A B C D$ with $A B | | D C$ intersect each other at the point $O .$ Using a similarity criterion for two triangles, show that $\frac{O A}{O C} = \frac{O B}{O D}$ .

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Solution

Given:
ABCD is a trapezium with $A B ∥ D C$ .
O is the point of intersection of two diagonals
To Prove:
$\frac{O A}{O C} = \frac{O B}{O D}$
Proof:
In $△ A O B$ and $△ D O C$
$∠ B A O = ∠ O C D$ (Alternate Angles)
$∠ A B O = ∠ O D C$ (Alternate Angles)
$∠ A O B = ∠ D O C$ (Vertically opposite angles)
$∴$ By AAA criterion of similarity, $△ A O B \sim △ D O C$
$∴$ $\frac{O A}{O C} = \frac{O B}{O D}$ (Corresponding Sides of Similar Triangles)

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