Question

The diagonals of a quadrilateral \[ABCD\] intersects each other at the point \[O\] such that \[\dfrac{{AO}}{{BO}} = \dfrac{{CO}}{{DO}}\]. Show that \[ABCD\] is a trapezium.

Answer 1

Hint: We will start by drawing the figure of quadrilateral \[ABCD\] such that \[\dfrac{{AO}}{{BO}} = \dfrac{{CO}}{{DO}}\]. Now draw a line parallel to one side of the quadrilateral say, AB. Then use the Basic proportionality theorem to prove that \[ABCD\] is a trapezium.

Complete step by step solution:
We are given a quadrilateral \[ABCD\] that intersects each other at the point \[O\] such that \[\dfrac{{AO}}{{BO}} = \dfrac{{CO}}{{DO}}\].
Here, we have to prove that \[ABCD\] is a trapezium using the condition \[\dfrac{{AO}}{{BO}} = \dfrac{{CO}}{{DO}}\].
According to Basic proportionality theorem, if we draw a line parallel to a side of the triangle so that it intersects with the other two sides, then the other two sides are divided in equal ratios.

If we draw a line \[EF\] parallel to one side \[AB\] of the quadrilateral, we can use the Basic proportionality theorem.
If \[EF||AB\] then \[EO||AB\] also.

As \[\dfrac{{AO}}{{BO}} = \dfrac{{CO}}{{DO}}\] equation(i) is given in the question.
By applying Basic proportionality theorem in \[\Delta ABD\] as this theorem deals with the properties of parallel lines in the triangles,
Thus, we have \[\dfrac{{AE}}{{ED}} = \dfrac{{DO}}{{OB}}\] equation(ii).

From equations (i) and (ii), we get,
\[\dfrac{{CO}}{{OA}} = \dfrac{{DO}}{{OB}}\] and \[\dfrac{{DO}}{{OB}} = \dfrac{{AE}}{{ED}}\]
\[
   \Rightarrow \dfrac{{CO}}{{OA}} = \dfrac{{DO}}{{OB}} = \dfrac{{AE}}{{ED}} \\
   \Rightarrow \dfrac{{CO}}{{OA}} = \dfrac{{AE}}{{ED}} \\
 \]

So, we have that in \[\Delta ACD\], the sides \[AC\], \[AD\] are divided in equal ratios. Thus, we can apply the converse of Basic proportionality theorem to get, side \[CD\] parallel to line \[EO\].
So, we have
\[
  AB||EO||CD \\
   \Rightarrow AB||CD \\
 \]

We know that if two sides in a quadrilateral are parallel to each other, then the quadrilateral is called a trapezium. Hence proved that \[ABCD\] is trapezium.

Note: The condition required to prove that a quadrilateral is a trapezium is to show that two opposite sides of the quadrilateral are parallel. Use the Basic proportionality theorem that states that, if we draw a line parallel to a side of the triangle so that it intersects with the other two sides, then the other two sides are divided in equal ratios. We have used this theorem because this theorem deals with the property of parallel lines in a triangle. And a quadrilateral can be easily divided in triangles.