prove that the line segment joining the midpoints of the diagonals of a Trapezium is parallel to the parallel sides and equal to the half their difference ..
Answers
Step-by-step explanation:
Let E and F are midpoints of the diagonals AC and BD of trapezium ABCD respectively.
Draw DE and produce it to meet AB at G
Consider △AEG and △CED
⇒ ∠AEG=∠CED [ Vertically opposite angles ]
⇒ AE=EC [ E is midpoint of AC ]
⇒ ∠ECD=∠EAG [ Alternate angles ]
⇒ △AEG≅△CED [ By SAA congruence rule ]
⇒ DE=EG ---- ( 1 ) [ CPCT ]
⇒ AG=CD ----- ( 2 )
In △DGB
E is the midpoint of DG [ From ( 1 ) ]
F is midpoint of BD
∴ EF∥GB
⇒ EF∥AB [ Since GB is part of AB ]
⇒ EF is parallel to AB and CD.
Also, EF=
2
1
GB
⇒ EF=
2
1
(AB−AG)
⇒ EF=
2
1
(AB−CD) [ From ( 2 ) ]
Answer:
Looking for a satisfying answer?? Here it is........
Step-by-step explanation:
Let E and F are midpoints of the diagonals AC and BD of trapezium ABCD respectively.
Draw DE and produce it to meet AB at G.
Consider DAEG and DCED
∠AEG = ∠CED (vertically opposite angles)
AE = EC (E is midpoint of AC)
∠ECD = ∠EAG (alternate angles)
ΔAEG ≅ ΔCED
⇒ DE = EG → (1)
And AG = CD → (2)
In ΔDGB
E is the midpoint of DG [From (1)]
F is the midpoint of BD
∴ EF is parallel to GB
⇒ EF is parallel to AB
⇒ EF is parallel to AB and CD
Also, EF = ½ GB
⇒EF = ½ (AB − AG)
⇒ EF = ½ (AB − CD) [From (2)]
Thus, proved that the line segment joining the midpoints of the diagonals of the trapezium is parallel to the parallel sides and equal to half their difference.
Hope this helps........ : )