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Triangles
EXERCISE 6.2
CBSE_NCERT_TEXTBOOK
Grade 10
Question:
ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O. Show that AO CO BO DO
Step-by-Step Solution
Key Concept: In a trapezium with one pair of opposite sides parallel, the intersecting diagonals form two pairs of similar triangles. Using the similarity of \(\triangle AOB \sim \triangle COD\) (or \(\triangle AOD \sim \triangle CBO\)) we obtain the proportionality \(\dfrac{AO}{CO}=\dfrac{BO}{DO}\). Cross‑multiplying gives the required relation \(AO\cdot DO = BO\cdot CO\), which, by the commutative property of multiplication, is equivalent to \(AO\cdot CO = BO\cdot DO\).
1. Identify the similar triangles\
Since \(AB \parallel DC\), the alternate interior angles give\
\[\angle ABO = \angle CDO \quad\text{and}\quad \angle BAO = \angle DCO.\]
Also, the vertical angles at the intersection point \(O\) give\
\[\angle AOB = \angle COD.\]
Hence \(\triangle AOB \sim \triangle COD\) (AA similarity).
2. Write the proportion from similarity\
From the correspondence \(A \leftrightarrow C,\; B \leftrightarrow D,\; O \leftrightarrow O\), we have\
\[\frac{AO}{CO}=\frac{AB}{CD}=\frac{BO}{DO}.\]
In particular,\
\[\frac{AO}{CO}=\frac{BO}{DO}.\]
3. Cross‑multiply\
\[AO\cdot DO = BO\cdot CO.\]
Since multiplication of real numbers is commutative, the equality can be written as\
\[AO\cdot CO = BO\cdot DO.\]
This is the required result.
4. Conclusion\
Thus, in a trapezium where the pair of opposite sides are parallel, the product of the segments of one diagonal equals the product of the segments of the other diagonal.
Remark: The same result can also be obtained by using the other pair of similar triangles \(\triangle AOD \sim \triangle CBO\). Both approaches lead to the same proportionality and hence the same product relation.
Since \(AB \parallel DC\), the alternate interior angles give\
\[\angle ABO = \angle CDO \quad\text{and}\quad \angle BAO = \angle DCO.\]
Also, the vertical angles at the intersection point \(O\) give\
\[\angle AOB = \angle COD.\]
Hence \(\triangle AOB \sim \triangle COD\) (AA similarity).
2. Write the proportion from similarity\
From the correspondence \(A \leftrightarrow C,\; B \leftrightarrow D,\; O \leftrightarrow O\), we have\
\[\frac{AO}{CO}=\frac{AB}{CD}=\frac{BO}{DO}.\]
In particular,\
\[\frac{AO}{CO}=\frac{BO}{DO}.\]
3. Cross‑multiply\
\[AO\cdot DO = BO\cdot CO.\]
Since multiplication of real numbers is commutative, the equality can be written as\
\[AO\cdot CO = BO\cdot DO.\]
This is the required result.
4. Conclusion\
Thus, in a trapezium where the pair of opposite sides are parallel, the product of the segments of one diagonal equals the product of the segments of the other diagonal.
Remark: The same result can also be obtained by using the other pair of similar triangles \(\triangle AOD \sim \triangle CBO\). Both approaches lead to the same proportionality and hence the same product relation.
Correct Answer: AO·CO = BO·DO (i.e., the product of the two parts of one diagonal equals the product of the two parts of the other diagonal).
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