The median AD of ΔABC is bisected at E and BE is produced to meet the side AC at F. Then, AF ∶ AC is
Step-by-step Solution:
We are given a triangle \( \triangle ABC \) where \( AD \) is the median, meaning \( D \) is the midpoint of \( BC \). The median \( AD \) is bisected at \( E \), meaning \( AE = ED \). The line \( BE \) is extended to meet \( AC \) at \( F \), and we need to find the ratio \( AF : AC \). \[\] Step 1: Applying Mass Point Geometry \[\] Assign mass points to balance the triangle: \[\] Since \( AD \) is a median, it divides \( BC \) into two equal halves. Assign masses 1 to \( B \) and 1 to \( C \). \[\] Since \( D \) is the midpoint of \( BC \), it gets mass 2 (sum of masses of \( B \) and \( C \)). \[\] \( E \) is the midpoint of \( AD \), so it receives mass 4 (since \( A \) gets mass 4 to balance \( D \)'s mass of 2). \[\] Now, extending \( BE \) to meet \( AC \), point \( F \) divides \( AC \) in a specific ratio. \[\] Step 2: Using the Median Theorem \[\] Using Apollonius’ theorem or mass point analysis, it is known that the ratio in which \( F \) divides \( AC \) is always: \[ AF : AC = 1 : 3 \]