A circle is inscribed in a triangle ABC. It touches the sides BC, CA, AB at D, E, F respectively. What is \angle EDF equal to?

  1. A. 90^{\circ}-A
  2. B. 90^{\circ}-\frac{(B+C)}{2}
  3. C. 90^{\circ}-2A
  4. D. 90^{\circ}-(\frac{A}{2})

Correct Answer: D. 90^{\circ}-(\frac{A}{2})

Explanation

In \triangle AEF, tangents AE and AF are equal, so \angle AEF = \angle AFE = \frac{180^{\circ} - A}{2} = 90^{\circ} - \frac{A}{2}. By the alternate segment theorem, the angle between the chord EF and tangent CA equals the angle subtended by EF at the circumference, giving \angle EDF = \angle AEF = 90^{\circ} - \frac{A}{2}.

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