In a parallelogram ABCD, E and F are the mid points of sides AB and CD respectively. Show that the line segments AF and EC trisect each other
Answers
ANSWER:-
Given:
In a parallelogram ABCD, E & F are the mid points of sides AB & CD respectively.
To find:
Show that the line segment AF & EC trisect each other.
Solution:
ABCD is a parallelogram.
So,
AB||CD & AE||FC
&
AB= CD [opposite sides of ||gm ABCD]
AE= FC
[since,E & F are mid-points of AB & CD]
In quadrilateral AECF,
One pair of opposite sides are equal & parallel.
Therefore,
AECF is a ||gm.
Then,
AE||FC
Again,
AB= CD [opposite sides of ||gm ABCD]
=) 1/2AB = 1/2CD
AE= FC
[E& F are mid-points of side AB & CD]
In quadrilateral AECF,
one pair of opposite sides are parallel & equal to each other.
Therefore,
AEFC is a parallelogram.
AF||EC. [opposite sides of a ||gm]
In ∆DQC,
F is the mid-point of side DC & FP||CQ
Therefore,
By using the Converse of midpoint theorem, it can be said that P is the mid-point of DQ.
So,
DP= PQ.......(1)
AB= CD
Similarly,
In ∆APB,
E is the mid-point of side AB & EQ||AP.
Therefore,
By using the converse of mid-point theorem, it can be said that Q is the mid-point of PB.
So,
PQ = QB........(2)
from equation (1) & (2), we get;
DP= PQ= BQ
Hence,
The line segments AF & EC trisect the diagonal BD.
Hope it helps ☺️
