TS EAMCET · Maths · Properties of Triangles
If the angular bisector of the angle A of the triangle ABC meets its circumcircle at E and the opposite side BC at D, then \(\mathrm{DE} \cos \frac{\mathrm{A}}{2}=\)
- A \(\frac{a^2}{2(b+c)}\)
- B \(\frac{b^2}{c+a}\)
- C \(\frac{a}{b+c}\)
- D \(\frac{2 a}{a+b+c}\)
Answer & Solution
Correct Answer
(A) \(\frac{a^2}{2(b+c)}\)
Step-by-step Solution
Detailed explanation
\( BD = \frac{ac}{b+c}, CD = \frac{ab}{b+c} \) \( AD \cdot DE = BD \cdot CD \) \( AD = \frac{2bc}{b+c} \cos \frac{A}{2} \) \( \left( \frac{2bc}{b+c} \cos \frac{A}{2} \right) DE = \frac{ac}{b+c} \cdot \frac{ab}{b+c} \)…
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