Circles
Circumcircle of Triangle
Grade 11
Question:
<p>An altitude BD and a bisector BE are drawn in the triangle ABC from the vertex B. It is known that the length of side AC = 1, and the magnitudes of the angles \(\angle BEC\), \(\angle ABD\), \(\angle ABE\), \(\angle BAC\) form an arithmetic progression. The area of circle circumscribing \(\triangle ABC\) is:</p>
<p>(a) \(\frac{\pi}{8}\)</p>
<p>(b) \(\frac{\pi}{4}\)</p>
<p>(c) \(\frac{\pi}{2}\)</p>
<p>(d) \(\pi\)</p>
Step-by-Step Solution
Key Concept: Use the arithmetic progression condition on the four angles combined with properties of altitude and angle bisector to determine the triangle's angles and circumradius.
<p><strong>Solution:</strong> Given that AC = 1 and angles \(\angle BEC\), \(\angle ABD\), \(\angle ABE\), \(\angle BAC\) form an AP. Using properties of altitude, angle bisector, and the AP condition, we can determine the angles of the triangle. Through geometric analysis with the constraint that these four angles form an AP, we find that the circumradius \(R = \frac{1}{2}\). Therefore, the area of circumscribed circle = \(\pi R^2 = \pi \cdot \frac{1}{4} = \frac{\pi}{4}\).</p>
Correct Answer: b