3
(1) In JABCD, diagonals AC and
BD intersect each other at point E.
then
complete
and the
of side
the following activity to prove
D ABCD is a trapezium.
Proof:
Draw line EF || side DC such that B-FC.
In ABDC,
Seg EL || side DC
(Construction)
BE
BF
also
ED
EC
(Given) ... (2)
срѕ.
. [From (1) and (2)) --- (3)
FC
rect
Now in AACD,
ircle
CE
AE
(By invertendo
seg EF || side AB
ted
(Given)
and seg EF | side
seg AB | side DC
. ABCD is a trapezium.
Answers
Answered by
0
Answer:
Here, ABCD is a quadrilateral with diagonals AC and BD intersect at point E.
⇒
EC
AE
=
ED
BE
[ Given ]
⇒
BE
AE
=
ED
EC
---- ( 1 )
In △ABE and △CDE,
⇒
BE
AE
=
ED
EC
] From ( 1 ) ]
⇒ ∠ABE=∠DEC [ Vertically opposites angles ]
∴ △ABE∼△CDE [ By SAS similarity theorem ]
⇒ ∠EDC=∠EBA [ Corresponding angles are equal. ]
∴ ∠BDC=∠ABD
Bu, this is a pair of alternate angles.
∴ AB∥DC
Thus, the quadrilateral ABCD is trapezium.
solution
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