in a parallelogram ...plz solve it
Answers
Answer:
HOPE IT HELPS PLS MARK AS BRAINLIEST
★ Given:-
- A parallelogram ABCD
- E and F are the mid-points of AB and CD
★ To Prove:-
- AF and EC trisects BD
★ Theoream Used:-
- Mid - Point theoream of triangle
★ Proof:-
As, E and F are the mid-points of side DC and AB
Since,
EB = ½AB-------eq. 1
CF = ½ DC------eq. 2
As,ABCD is a parallelogram
Hence,
AB = CD
AB || CD
So, ½AB = ½CD
AE = FC
So, from equation 1 and 2
As,
AE = FC and AE || FC
∴ AEFC is a parallelogram.
FA || CE
As, FP is a part of FA and CQ is a part of CE
FP || CQ ------ eq.3
Since,
The segment drawn through the mid-point of one side of a triangle and parallel to the other side bisects the third side.
In ∆ DCQ,
F is the mid-point of CD and FP || CQ.
∴ P is the midpoint of DQ.
DP = PQ ------ eq.4
Similarly,
In ∆ ABP, Q is the mid point of BP.
By mid-point theoream
BQ = PQ ----- eq.5
From eq. 4 and 5
DP = PQ = BQ ----- eq.6
Diagonal BD
= BQ + PQ + DP
= BQ + BQ + BQ (From eq.6)
BD = 3BQ
BQ = ⅓BD------ eq.7
Similarly
From eq. 6 and 7
DP = PQ = BQ = ⅓BD
Therefore, P and trisects line BD.
Hence , it is proved that AF and EC trisects the diagonal BD.
