The diagonals of a quadrilateral ABCD intersect each other at the point O such that
АО/BO
CO/DO
Show that ABCD is a trapezium.
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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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