1.

Prove diagonals of a quadrilateral ABCD intersect at O. AO/BO=CO/DO​

Answer»

ANSWER:

Given: △ABCD is a trapezium with AB || CD and the DIAGONALS AC and BD

intersect at ‘0'.

To prove: OA/OC = OB/OD

Proof:

In the figure consider the triangle OAB and OCD

∠DOC = ∠AOB (Vertically opposite ANGLES are equal)

since AB || DC,

∠DOC = ∠OAB (Alternate angles are equal)

∴ By AA corollary of similar triangles.

∴ △OAB ~ △OCB When the two triangle are similar, the side are proportionally.

⇒ OA/OC = OB/OD

Hence proved.


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