The diagonals of a quadrilateral ABCD intersect each other
at the point O
such that
AO
BO
=
CO
DO
show that ABCD is a trapezium
Proof :
BO
AO
=
CO
DO
∴
CO
AO
=
BO
DO
…(i)
In Δ AOB and ΔCOD,
D
A
B
C
[By SAS criterion]
∴
ABCD is a trapezium.
CO
AO
=
BO
DO
[From (i)]
∠AOB
=
∠COD
[Vertically Opposite
angles]
ΔAOB ~ ΔCOD
∠ABO
=
∠CDO
[c.a.s.t.]
∠ABD
=
∠CDB
AB ll DC
Hint : To prove
AB II BC
EX.6.2 Q.10
O
∴
∴
But, these are a pair of alternate
interior angles on transversal BD.