Vector Algebra
Vectors
star_batch_jee_advanced_2025
Grade 12
Question:
$ABC$ is a triangle. $AD, AD'$ are internal and external bisectors of angle $A$, meeting $BC$ at $D$ and $D'$ respectively. $A'$ is the mid-point of $DD'$ and $B', C'$ are similar points on $CA$ and $AB$. Then $A', B', C'$:
Lie on a plane
Form an equilateral triangle
Form an isosceles triangle
None of these
Step-by-Step Solution
Key Concept: The midpoints of segments between internal and external angle bisector feet on each side form a coplanar configuration due to the harmonic division properties of angle bisectors.
By the angle bisector theorem, $D$ divides $BC$ internally in ratio $AB:AC$, while $D'$ divides $BC$ externally in the same ratio. The midpoint $A'$ of $DD'$ lies on the perpendicular from $A$ to $BC$. By symmetry, points $B'$ (midpoint of $EE'$ on $CA$) and $C'$ (midpoint of $FF'$ on $AB$) are similarly constructed. These three points $A'$, $B'$, $C'$ are the midpoints of segments connecting internal and external angle bisector feet, which form the Simson-like configuration. Using properties of angle bisectors and midpoints, these three points always lie on a single plane (in fact, they form the nine-point circle or lie collinearly/coplanarly depending on triangle orientation), satisfying a fundamental geometric relationship.
Correct Answer: 1