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

Step-by-Step Solution:

1. Understand the Given Information:

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

2. To Prove: $$\frac{AO}{OC} = \frac{BO}{OD}$|

3. Consider the Triangles Formed: In $\triangle AOB$ and $\triangle COD$:

   • Angle $\angle AOB$ is vertically opposite to angle $\angle COD$: $$\angle AOB = \angle COD \quad \text{(Vertically opposite angles)}$$
   • Since $AB \parallel DC$ and $BD$ is a transversal line, the alternate interior angles are equal: $$\angle ABO = \angle CDO \quad \text{(Alternate interior angles)}$$
   • Similarly, considering $AC$ as a transversal line between $AB \parallel DC$: $$\angle BAO = \angle DCO \quad \text{(Alternate interior angles)}$$

4. Apply Similarity Criterion: By using the AAA (Angle-Angle-Angle) similarity criterion, all three corresponding angles of $\triangle AOB$ and $\triangle COD$ are equal. Therefore, the two triangles are similar: $$\triangle AOB \sim \triangle COD$$

5. Use Properties of Similar Triangles: We know that the corresponding sides of similar triangles are in proportion (proportional): $$\frac{AO}{CO} = \frac{BO}{DO} = \frac{AB}{CD}$|

6. Final Rearrangement: Taking the first two parts of the equality: $$\frac{AO}{OC} = \frac{BO}{OD}$| Hence proved.