# Ex.6.2 Q10 Triangles Solution - NCERT Maths Class 10

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## Question

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

Reasoning:

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.

Steps:

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 $$(2)$$

\begin{align}\frac{OB}{OD}=\frac{BE}{CE}\,\,\,\end{align}

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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