Question

Q. Diagonals AC and BD of a trapezium ABCD with AB||DC intersect each other at the point O. Using a similarity criterion for two triangles show that OA/OC= OB/OD.(No spam)
.
.
thank you so much ❤️​

Answer 1

Answer:

ABCD is a trapezium with AB∥CD and diagonals AB and CD intersecting at O.

⇒ In △OAB and △OCD

⇒ ∠AOB=∠DOC [ Vertically opposite angles ]

⇒ ∠ABO=∠CDO [ Alternate angles ]

⇒ ∠BAO=∠OCD [ Alternate angles ]

∴ △OAB∼△OCD [ AAA similarity ]

We know that if triangles are similar, their corresponding sides are in proportion.

⇒

OC

OA

=

OD

OB

[henceproved]

Step-by-step explanation:

hope this will help to u

please mark me brainliest

Answer 2

Answer:

ABCD is a trapezium with AB∥CD and diagonals AB and CD intersecting at O.

⇒ In △OAB and △OCD

⇒ ∠AOB=∠DOC [ Vertically opposite angles ]

⇒ ∠ABO=∠CDO [ Alternate angles ]

⇒ ∠BAO=∠OCD [ Alternate angles ]

∴ △OAB∼△OCD [ AAA similarity ]

We know that if triangles are similar, their corresponding sides are in proportion.

⇒ OC. OD

OO = OB

[henceproved]