Anonymous
Jan 18
ABCD is a trapezium in which AB P DC and its diagonals intersect each other at the point 'O'. Show that AO CO B DO = . ?
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ChatGPT
Jan 18
To prove that AO/CO = BO/DO, we can use the property of similar triangles.
Let's consider triangle AOB and triangle COD.
Since AB || CD (as ABCD is a trapezium), we have ∠AOB = ∠COD (alternate angles).
Also, ∠ABO = ∠CDO (corresponding angles).
Therefore, by AA similarity, triangle AOB ~ triangle COD.
By the property of similar triangles, we have:
AO/CO = BO/DO.
Hence, AO/CO = BO/DO.
Let's consider triangle AOB and triangle COD.
Since AB || CD (as ABCD is a trapezium), we have ∠AOB = ∠COD (alternate angles).
Also, ∠ABO = ∠CDO (corresponding angles).
Therefore, by AA similarity, triangle AOB ~ triangle COD.
By the property of similar triangles, we have:
AO/CO = BO/DO.
Hence, AO/CO = BO/DO.
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