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Home » Diagonals AC and BD of a trapezium ABCD with AB intersect each other at the point Using a similarity criterion for two triangles, show that
Diagonals AC and BD of a trapezium ABCD with AB \| \mathrm{DC} intersect each other at the point \mathrm{O} . Using a similarity criterion for two triangles, show that \mathbf{A O} / \mathbf{O C}=\mathbf{O B} / \mathbf{O D}
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Diagonals AC and BD of a trapezium ABCD with AB \| \mathrm{DC} intersect each other at the point \mathrm{O} . Using a similarity criterion for two triangles, show that \mathbf{A O} / \mathbf{O C}=\mathbf{O B} / \mathbf{O D}

Solution:
In \Delta \mathrm{DOC} and \triangle \mathrm{BOA}
\mathrm{AB} \| \mathrm{CD}, thus alternate interior angles will be equal, \therefore \angle \mathrm{CDO}=\angle \mathrm{ABO}
Similarly,

    \[\angle \mathrm{DCO}=\angle \mathrm{BAO}\]

Also, for the two triangles \Delta \mathrm{DOC} and \Delta \mathrm{BOA}, vertically opposite angles will be equal; \therefore \angle \mathrm{DOC}=\angle \mathrm{BOA}
Hence, by AAA similarity criterion,

    \[\Delta \mathrm{DOC} \sim \triangle \mathrm{BOA}\]

Thus, the corresponding sides are proportional.

    \[\begin{array}{l} \mathrm{DO} / \mathrm{BO}=\mathrm{OC} / \mathrm{OA} \\ \Rightarrow \mathrm{OA} / \mathrm{OC}=\mathrm{OB} / \mathrm{OD} \end{array}\]

Hence, proved.

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