Ex.6.2 Q10 Triangles Solution - NCERT Maths Class 10

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The diagonals of a quadrilateral \(ABCD\) intersect each other at the point \(‘O’\) such that \(\begin{align}\frac{AO}{BO}=\frac{CO}{DO}\end{align}\). Show that \(ABCD\) is a trapezium.

Text Solution



As we know if a line divides any two sides of a triangle in the same ratio,then the line is parallel to the third side.


In quadrilateral \(ABCD\)

Diagonals \(AC, BD\) intersect at \(‘O’\)

Draw \( OE||AB\)

\(\begin{align}& \text{In}\,\,\,\Delta \,ABC \\ &\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,OE||AB \\ \end{align}\)

\[\Rightarrow \frac{{OA}}{OC} = \frac{{BE}}{CE}  \,\,(BPT)................(1)\]

But \(\begin{align}\frac{OA}{OB}=\frac{OC}{OD}\,\,\left( \rm{given} \right)\end{align}\)

\[\begin{align}\Rightarrow \frac{{OA}}{OC}=\frac{{OB}}{{OD}}..............\left( 2 \right)\end{align}\]

From \((1)\) and \((2)\)


In \(\Delta BCD\)

\[\begin{align}\rm \frac{OB}{OD}&=\frac{BE}{CE} \\ \rm OE&\,\,||CD \\ \rm OE& \,\,||AB\,\,\text{and}\,\,\rm OE||CD \\ \Rightarrow \rm AB& \,\,||CD \\ &\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\Rightarrow \rm ABCD\,\,\text{is a trapezium} \end{align}\]

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