NCERT Exemplar Solutions
Class 10 - Mathematics - CHAPTER 9: Circles - Exercise 9.2Question 7
Class 10 - Mathematics - CHAPTER 9: Circles - Exercise 9.2
Question. 7
In Fig. 9.14, common tangents \(AB\) and \(CD\) to two circles intersect at \(E\). Prove that \(AB=CD\).
Answer:
\(AB=CD\)
Detailed Answer with Explanation:
Step 1: Recall a basic property of tangents: From a point outside a circle, the two tangents drawn to that circle are always equal in length.
Step 2: Here, point \(E\) is a common external point. - For the left circle: the two tangents are \(EA\) and \(EC\). So, \(EA = EC\).
Step 3: For the right circle: the two tangents from \(E\) are \(EB\) and \(ED\). So, \(EB = ED\).
Step 4: Now look at line segment \(AB\). It can be written as: \(AB = EA - EB\).
Step 5: Similarly, line segment \(CD\) can be written as: \(CD = EC - ED\).
Step 6: Since we already know that \(EA = EC\) and \(EB = ED\), we can replace them: \(AB = EA - EB = EC - ED = CD\).
Final Result: Therefore, \(AB = CD\).