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Question 3Dia...

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Question 3
Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at the point O. Using a similarity criterion for two triangles, show that $\frac{A O}{O C} = \frac{O B}{O D}$ .

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Solution

In $Δ D O C$ and $Δ B O A$ ,
$∠ C D O = ∠ A B O$ [Alternate interior angles as AB || CD]
$∠ D C O = ∠ B A O$ [Alternate interior angles as AB || CD]
$∠ D O C = ∠ B O A$ [Vertically opposite angles]
$∴ Δ D O C \sim Δ B O A$ [AAA similarity criterion]
$∴ \frac{D O}{B O} = \frac{O C}{O A}$ [Corresponding sides are proportional]
$\Rightarrow \frac{O A}{O C} = \frac{O B}{O D}$

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