MP Board · Class 10 · Mathematics · Trianglesएक समलंब चतुर्भुज $ABCD$ जिसमें $AB \parallel DC$ है, के विकर्ण $AC$ और $BD$ परस्पर बिंदु $O$ पर प्रतिच्छेद करते हैं। त्रिभुजों की समरूपता की कसौटी का प्रयोग करते हुए दर्शाइए कि $\frac{OA}{OC} = \frac{OB}{OD}$ है।

Step-by-Step Solution:

1. Understand the Given Information:

• $ABCD$ is a trapezium in which $AB \parallel DC$.
• Diagonals $AC$ and $BD$ intersect each other at point $O$.

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

3. Consider Triangles $AOB$ and $COD$:\nLet us look at $\triangle AOB$ and $\triangle COD$.

• Since $AB \parallel DC$ and $BD$ acts as a transversal line, the alternate interior angles are equal: $$\angle OAB = \angle OCD \quad \text{--- (Alternate interior angles)}$$
• Similarly, taking $AC$ as the transversal line for parallel lines $AB$ and $DC$, we get: $$\angle OBA = \angle ODC \quad \text{--- (Alternate interior angles)}$$
• Also, the vertically opposite angles formed at the intersection point $O$ are equal: $$\angle AOB = \angle COD \quad \text{--- (Vertically opposite angles)}$$

4. Apply Similarity Criterion:\nBy AAA (Angle-Angle-Angle) similarity criterion, since all three corresponding angles of $\triangle AOB$ and $\triangle COD$ are equal: $$\triangle AOB \sim \triangle COD$$

5. Establish Ratio of Corresponding Sides:\nWe know that if two triangles are similar, then the ratio of their corresponding sides is equal (Proportionality of similar triangles): $$\frac{OA}{OC} = \frac{OB}{OD} = \frac{AB}{CD}$| \nTaking the first two parts of the equality: $$\frac{OA}{OC} = \frac{OB}{OD}$| \nHence proved.