Question

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.​

Answer 1

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