In a trapezium
\(
ABCD\) where
\(
AB\) is parallel to
\(
DC\), the diagonals intersect at
\(O\). Establish that
\(\frac{AO}{BO}=\frac{CO}{DO}\)
Proof:
In \(\Delta ADC\),
\(MO||\)
By
So, \(\frac{AM}{DM} =\) - - - - - (1)
Similarly, in \(\Delta DBA\)
\(MO||\)
\(\frac{AM}{DM} =\) - - - - - (2)
From (1) and (2) we proved the result.
Answer variants:
\(\frac{BO}{DO}\)
\(DC\)
\(AB\)
\( \frac{AO}{CO}\)
\( \frac{AO}{DO}\)
\(\frac{BO}{CO}\)