Question

Diagonals AC and BD of a trapezium ABCD
with AB || DC intersect each other at the
point O. Using a similarity criterion for two
OA OB
triangles, show that oc - OD​

Answer 1

Answer:

here is your answer......

Answer 2

In ΔDOC and ΔBOA,

AB || CD, thus alternate interior angles will be equal,

∴∠CDO = ∠ABO

Similarly,

∠DCO = ∠BAO

Also, for the two triangles ΔDOC and ΔBOA, vertically opposite angles will be equal;

∴∠DOC = ∠BOA

Hence, by AAA similarity criterion,

ΔDOC ~ ΔBOA

Thus, the corresponding sides are proportional.

DO/BO = OC/OA

⇒OA/OC = OB/OD

Hence, proved.