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Triangles
EXERCISE 6.3
CBSE_NCERT_TEXTBOOK
Grade 10
Question:
Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at the point O. Using a similarity criterion for two , show that OA OB OC OD Fig. 6.35 96
Step-by-Step Solution
Key Concept: In a trapezium the pair of opposite sides are parallel. By using the parallelism, we can establish equal corresponding angles, which together with the vertical angle give two pairs of similar triangles formed by the intersecting diagonals. From the similarity we obtain the proportionality of the corresponding sides, and on cross‑multiplication we get the required product relation.
1. Identify the triangles\
The intersecting diagonals AC and BD create four small triangles: \(\triangle AOB, \triangle BOC, \triangle COD, \triangle DOA\).\
We will prove that \(\triangle AOB \sim \triangle COD\) and \(\triangle AOD \sim \triangle CBO\).\
2. Angles in \(\triangle AOB\) and \(\triangle COD\)\
- \(\angle AOB\) and \(\angle COD\) are vertical angles, therefore \(\angle AOB = \angle COD\).\
- Since \(AB \parallel DC\), the angle formed by a transversal BD with the two parallel lines are equal: \(\angle ABO = \angle CDO\).\
Hence two angles of \(\triangle AOB\) are equal to two angles of \(\triangle COD\); consequently, \(\triangle AOB \sim \triangle COD\) (AA similarity).\
3. Corresponding sides from the similarity\
From \(\triangle AOB \sim \triangle COD\) we have\
$$\frac{OA}{OC}=\frac{OB}{OD}=\frac{AB}{DC}\tag{1}$$\
Rearranging the first equality gives\
$$OA\cdot OD = OB\cdot OC.\tag{2}$$\
This is the required relation.
4. (Optional) Similarity of the other pair\
Using the same reasoning with the other pair of triangles, \(\triangle AOD\) and \(\triangle CBO\), we obtain the same proportion, confirming the result.
5. Conclusion\
Hence, for the intersecting diagonals of a trapezium, the product of the segments of one diagonal equals the product of the segments of the other diagonal:
$$\boxed{\;OA\cdot OD = OB\cdot OC\;}. $$
The intersecting diagonals AC and BD create four small triangles: \(\triangle AOB, \triangle BOC, \triangle COD, \triangle DOA\).\
We will prove that \(\triangle AOB \sim \triangle COD\) and \(\triangle AOD \sim \triangle CBO\).\
2. Angles in \(\triangle AOB\) and \(\triangle COD\)\
- \(\angle AOB\) and \(\angle COD\) are vertical angles, therefore \(\angle AOB = \angle COD\).\
- Since \(AB \parallel DC\), the angle formed by a transversal BD with the two parallel lines are equal: \(\angle ABO = \angle CDO\).\
Hence two angles of \(\triangle AOB\) are equal to two angles of \(\triangle COD\); consequently, \(\triangle AOB \sim \triangle COD\) (AA similarity).\
3. Corresponding sides from the similarity\
From \(\triangle AOB \sim \triangle COD\) we have\
$$\frac{OA}{OC}=\frac{OB}{OD}=\frac{AB}{DC}\tag{1}$$\
Rearranging the first equality gives\
$$OA\cdot OD = OB\cdot OC.\tag{2}$$\
This is the required relation.
4. (Optional) Similarity of the other pair\
Using the same reasoning with the other pair of triangles, \(\triangle AOD\) and \(\triangle CBO\), we obtain the same proportion, confirming the result.
5. Conclusion\
Hence, for the intersecting diagonals of a trapezium, the product of the segments of one diagonal equals the product of the segments of the other diagonal:
$$\boxed{\;OA\cdot OD = OB\cdot OC\;}. $$
Correct Answer: OA·OD = OB·OC
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