Circles
Angle Bisector and Circumcircle
Grade 11
Question:
<p>Let the bisector of ∠A of △ABC meets BC in D and the circumcircle of △ABC in E. Analyze: AD is less than the G.M. (Geometric Mean) of AB and AC.</p>
<p>(A) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1</p>
<p>(B) Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1</p>
<p>(C) Statement-1 is true, Statement-2 is false</p>
<p>(D) Statement-1 is false, Statement-2 is true</p>
Step-by-Step Solution
Key Concept: The angle bisector theorem combined with similarity of triangles ABD and AEC shows that AD is less than the geometric mean of AB and AC.
<p><strong>Analysis:</strong> Since △ABD ∼ △AEC (by angle bisector properties), we have \(\frac{AB}{AE} = \frac{AD}{AC}\), so \(AD = \frac{AB \cdot AC}{AE}\). Since E is on the circumcircle beyond D, \(AE > AD\), thus \(AD < \sqrt{AB \cdot AC}\) (G.M.). Statement-2 establishes the similarity which justifies Statement-1.</p>
Correct Answer: A