In a parallelogram ABCD E and F are the midpoints of sides ab and CD respectively show that the line segments AF and EC trisect the diagonal BD
Answers
Parallelogram :
A quadrilateral in which both pairs of opposite sides are parallel is called a parallelogram.
A quadrilateral is a parallelogram if
i)Its opposite sides are equal
ii) its opposite angles are equal
iii) diagonals bisect each other
iv) a pair of opposite sides is equal and parallel.
Converse of mid point theorem:
The line drawn through the midpoint of one side of a triangle, parallel to another side bisect the third side.
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Given,
ABCD is a parallelogram. E and F are the mid-points of sides AB and CD respectively.
To show: line segments AF and EC trisect the diagonal BD.
Proof,
ABCD is a parallelogram
Therefore, AB || CD
also, AE || FC
Now,
AB = CD
(Opposite sides of parallelogram ABCD)
1/2 AB = 1/2 CD
AE = FC
(E and F are midpoints of side AB and CD)
Since a pair of opposite sides of a quadrilateral AECF is equal and parallel.
so,AECF is a parallelogram
Then, AF||EC,
AP||EQ & FP||CQ
(Since opposite sides of a parallelogram are parallel)
Now,
In ΔDQC,
F is mid point of side DC & FP || CQ
(as AF || EC).
So,P is the mid-point of DQ
(by Converse of mid-point theorem)
DP = PQ — (i)
Similarly,
In APB,
E is mid point of side AB and EQ || AP
(as AF || EC).
So,Qis the mid-point of PB
(by Converse of mid-point theorem)
PQ = QB — (ii)
From equations (i) and (ii),
DP = PQ = BQ
Hence, the line segments AF and EC trisect the diagonal BD.
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TANU81 Ace
Hi friend ✋
GIVEN:-E and F are the mid point of sides AB and CD of the parallelogram ABCD whose diagonal is BD.
TO PROVE:-BQ=QP=PD
PROOF:-ABCD is parellogram(given)
AB||DC and AB||DC(opposite side is ||gm)
E is the mid point of AB
AE=1/2AB............((1))
F is the mid point CD
CF=1/2CD
CF=1/2AB ------(2)
From 1 and 2
AE =CF.
Also AE || CF
Thus, a pair of opposite sides os a quadrilateral AECF are parallel and equal .
Quadrilateral,AECF id a parellogram
=EC || AF
=EQ || AP and QC || MF
.
In triangle BMA ,E. is the mid point of BA. given
EQ || AP. proved
BQ=LP
Similar by taking triangle CLD ,we can prove that
DP=QP
From 3 and 4 we get
BQ=QP=PD
Hence, AF and CE trisect the diagonal AC.
Hope help ....