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Base BC of triangle ABC is fixed and opp...

Base BC of triangle ABC is fixed and opposite vertex A moves in such a way that `tan.(B)/(2)tan.(C)/(2)` is constant. Prove that locus of vertex A is ellipse.

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To prove that the locus of vertex A in triangle ABC is an ellipse when the base BC is fixed and the condition \( \tan\left(\frac{B}{2}\right) \tan\left(\frac{C}{2}\right) = k \) (a constant) holds, we can follow these steps: ### Step-by-Step Solution: 1. **Understanding the Given Condition**: We know that the base BC is fixed, and we denote the lengths of sides opposite to vertices A, B, and C as a, b, and c respectively. The angles at vertices B and C are denoted as B and C. 2. **Using the Half-Angle Tangent Formula**: ...
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