LIVE Course for free

Rated by 1 million+ students
Get app now
0 votes
in Quadrilaterals by (88.7k points)
closed by
E and F are the mid-points of the sides AB and CD of a parallelogram ABCD. Prove that the line segment AF and CE trisects BD in three equal parts.

1 Answer

0 votes
by (97.1k points)
selected by
Best answer
ABCD is a parallelogram
`therefore" "AB=DCand AB"||"DC`
`implies" "2AE=2CFand AE"||"CF`
`implies" "AE=CFand AE"||"CF`
`impliessquareAECF` is a parallelogram
`therefore" "AF"||"EC`
In `DeltaDCM,`
F is the mid-point of DC
and `FN"||"CM`
`therefore` N is the mid-point of DM.
`implies" "DN=MN" "...(1)`
In `DeltaBAN,`
E is the mid-point of AB.
`and" "EM"||"AN`
`therefore` M is the mid-point of BN.
`implies" "BM=MN" "...(2)`
From eqs, (1) and (2)
`" "BM=MN=ND`
`implies` AF and CE, divides the diagonal BD into three equal parts.

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.