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In the previous problem, if O is the poi...

In the previous problem, if O is the point on AD and `OO_(1)` is the perpendicular from the centre of the circle, `O_(-1), then OA//OD` will be

A

`1`

B

`gt1`

C

`lt1`

D

depands upon `vec(B)` and charge

Text Solution

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The correct Answer is:
To solve the problem step by step, let's analyze the information given and apply the principles of geometry and physics as discussed in the video transcript. ### Step-by-Step Solution: 1. **Understanding the Setup**: - We have a point O on line AD. - There is a perpendicular line drawn from the center of the circle (point O1) to point O, denoted as OO1. - We need to find the ratio OA/OD. 2. **Identifying the Triangles**: - The problem mentions two triangles: triangle OOO1 (formed by points O, O1, and the center of the circle) and triangle AOO1 (formed by points A, O, and O1). - Since OO1 is perpendicular to AD, both triangles share the common side OO1. 3. **Congruent Triangles**: - The triangles OOO1 and AOO1 are congruent. This can be established because: - They share the side OO1. - The angles at O (angle AOO1 and angle OOO1) are equal (both are right angles). - The angles at O1 are also equal since they are subtended by the same line (the radius of the circle). - By the criteria of congruence (Angle-Side-Angle), we conclude that these triangles are congruent. 4. **Establishing the Ratio**: - Since the triangles are congruent, the corresponding sides are equal. Therefore: - OA = OD - This leads us to conclude that the ratio of OA to OD is: \[ \frac{OA}{OD} = 1 \] 5. **Final Answer**: - Thus, the final answer for the ratio OA/OD is: \[ \boxed{1} \]

To solve the problem step by step, let's analyze the information given and apply the principles of geometry and physics as discussed in the video transcript. ### Step-by-Step Solution: 1. **Understanding the Setup**: - We have a point O on line AD. - There is a perpendicular line drawn from the center of the circle (point O1) to point O, denoted as OO1. - We need to find the ratio OA/OD. ...
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