Maths · Circle, conic sections: A standard form of equations of a circle, the general form of the equation of a circle, its radius and centre
In the figure, is the point of intersection of two chords and such that , then t
In the figure, \( \boldsymbol{O} \) is the point of intersection of two chords \( A B \) and \( C D \) such that \( O B=O D \), then triangles \( O A C \) and \( O D B \) are:
- A. Equilateral but not similar
- B. Isosceles but not similar
- C. Equilateral and similar
- D. Isosceles and similar
Step-by-step solution
Given chords AB and CD intersect at O, with OB = OD. By the intersecting chords theorem, OA × OB = OC × OD. Since OB = OD, we get OA = OC. Thus, triangle OAC has OA = OC (isosceles), and triangle ODB has OB = OD (isosceles). Also, vertically opposite angles ∠AOC = ∠BOD. Therefore, by SAS similarity (sides proportional and included angle equal), triangles OAC and ODB are similar.
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