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Q3:
Diagonals $AC$ and $BD$ of a trapezium $ABCD$ with $AB \parallel DC$ intersect each other at the point $O$. Using a similarity criterion for two triangles, show that $\frac{OA}{OC} = \frac{OB}{OD}$.

Solution :

Given: A trapezium $ABCD$ in which $AB \parallel DC$. The diagonals $AC$ and $BD$ intersect each other at point $O$.

To Prove: $\frac{OA}{OC} = \frac{OB}{OD}$

A B C D O

Step 1: Identifying the Triangles
Consider $\triangle AOB$ and $\triangle COD$. We aim to prove that these two triangles are similar using the Angle-Angle (AA) similarity criterion.

Step 2: Establishing Equality of Angles
Since $AB \parallel DC$ and $AC$ acts as a transversal, the alternate interior angles are equal. Therefore:
$\angle OAB = \angle OCD$ [Alternate interior angles are equal when lines are parallel]
Similarly, since $AB \parallel DC$ and $BD$ acts as a transversal:
$\angle OBA = \angle ODC$ [Alternate interior angles are equal when lines are parallel]

Step 3: Considering Vertically Opposite Angles
The diagonals $AC$ and $BD$ intersect at $O$. Thus:
$\angle AOB = \angle COD$ [Vertically opposite angles are equal]

Step 4: Applying the Similarity Criterion
In $\triangle AOB$ and $\triangle COD$:
1. $\angle OAB = \angle OCD$ (Proved in Step 2)
2. $\angle OBA = \angle ODC$ (Proved in Step 2)
3. $\angle AOB = \angle COD$ (Proved in Step 3)
By the $AAA$ (Angle-Angle-Angle) similarity criterion, which simplifies to $AA$ similarity:
$\triangle AOB \sim \triangle COD$

Step 5: Establishing the Ratio of Corresponding Sides
Since the triangles are similar, the ratios of their corresponding sides must be equal:
$\frac{OA}{OC} = \frac{OB}{OD} = \frac{AB}{CD}$ [Corresponding parts of similar triangles are proportional (CPST)]

Step 6: Conclusion
From the proportionality established in Step 5, we extract the required equality:
$\frac{OA}{OC} = \frac{OB}{OD}$

Final Answer: Hence, it is proved that $\frac{OA}{OC} = \frac{OB}{OD}$.


More Questions from Class 10 Mathematics Triangles EXERCISE 6.3


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