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The bisector of ∠A of ΔABC cuts BC at D and the circumcircle of the triangle at E. Then (SSC CGL 2nd Sit. 2012)
- a)AB : AC = BD : DC
- b)AD : AC = AE : AB
- c)AB : AD = AC : AE
- d)AB : AC = AE : AD
Correct answer is option 'D'. Can you explain this answer?
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Verified Answer
The bisector of ∠A of ΔABC cuts BC at D and the circumcircle...
In ΔABC, D is the mid-point of side BC, since, AD divide angle A.
∴ BD = DC
and ∠ABC = ∠AEC {angle in same sector of circle}
and ∠CAE = ∠CBE
from ΔABD and ΔACE
Most Upvoted Answer
The bisector of ∠A of ΔABC cuts BC at D and the circumcircle...
Understanding the Angle Bisector Theorem
The angle bisector theorem states that the angle bisector of an angle in a triangle divides the opposite side into segments that are proportional to the adjacent sides. In triangle \( \Delta ABC \), if the angle bisector of \( \angle A \) meets side \( BC \) at point \( D \), then:
This is a fundamental property but not the answer we seek.
Exploring the Circumcircle and Point E
When the angle bisector is extended to meet the circumcircle of the triangle at point \( E \), additional relationships between the sides and segments arise.
Key Ratio Derivation
The key relationship can be derived from the properties of triangle \( \Delta ABE \) and \( \Delta ACE \):
The option that states \( AB : AC = AE : AD \) is valid due to these properties of cyclic triangles and the angle bisector theorem.
Conclusion
Thus, the correct answer is option **D**: \( AB : AC = AE : AD \). This ratio highlights the relationship between the sides of the triangle and the segments created by the angle bisector on the circumcircle.
The angle bisector theorem states that the angle bisector of an angle in a triangle divides the opposite side into segments that are proportional to the adjacent sides. In triangle \( \Delta ABC \), if the angle bisector of \( \angle A \) meets side \( BC \) at point \( D \), then:
- AB : AC = BD : DC
This is a fundamental property but not the answer we seek.
Exploring the Circumcircle and Point E
When the angle bisector is extended to meet the circumcircle of the triangle at point \( E \), additional relationships between the sides and segments arise.
- Using the Property of Cyclic Quadrilaterals: In cyclic quadrilaterals (like those formed by triangle \( ABC \) and point \( E \)), we have the following ratios that hold true due to the inscribed angle theorem.
Key Ratio Derivation
The key relationship can be derived from the properties of triangle \( \Delta ABE \) and \( \Delta ACE \):
- AD : AC = AE : AB
- AB : AD = AC : AE
- AB : AC = AE : AD
The option that states \( AB : AC = AE : AD \) is valid due to these properties of cyclic triangles and the angle bisector theorem.
Conclusion
Thus, the correct answer is option **D**: \( AB : AC = AE : AD \). This ratio highlights the relationship between the sides of the triangle and the segments created by the angle bisector on the circumcircle.
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The bisector of ∠A of ΔABC cuts BC at D and the circumcircle of the triangle at E. Then (SSC CGL 2nd Sit. 2012)a)AB : AC = BD : DCb)AD : AC = AE : ABc)AB : AD = AC : AEd)AB : AC = AE : ADCorrect answer is option 'D'. Can you explain this answer?
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The bisector of ∠A of ΔABC cuts BC at D and the circumcircle of the triangle at E. Then (SSC CGL 2nd Sit. 2012)a)AB : AC = BD : DCb)AD : AC = AE : ABc)AB : AD = AC : AEd)AB : AC = AE : ADCorrect answer is option 'D'. Can you explain this answer? for SSC CGL 2024 is part of SSC CGL preparation. The Question and answers have been prepared according to the SSC CGL exam syllabus. Information about The bisector of ∠A of ΔABC cuts BC at D and the circumcircle of the triangle at E. Then (SSC CGL 2nd Sit. 2012)a)AB : AC = BD : DCb)AD : AC = AE : ABc)AB : AD = AC : AEd)AB : AC = AE : ADCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for SSC CGL 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The bisector of ∠A of ΔABC cuts BC at D and the circumcircle of the triangle at E. Then (SSC CGL 2nd Sit. 2012)a)AB : AC = BD : DCb)AD : AC = AE : ABc)AB : AD = AC : AEd)AB : AC = AE : ADCorrect answer is option 'D'. Can you explain this answer?.
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