# In trapezium ABCD, \[M \in \overline {AD} ,N \in \overline {BC} \] are the points such \[\dfrac{{AM}}{{MD}} = \dfrac{{BN}}{{NC}} = \dfrac{2}{3}\]. The diagonal \[\overline {AC} \] intersects \[\overline {MN} \] at\[O\]. Then find the value of \[\dfrac{{AO}}{{AC}}\].

Last updated date: 24th Mar 2023

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Hint: A trapezium is a 2D shape which falls under the category of quadrilaterals. A trapezium has two parallel sides and two non-parallel sides. Using this information first draw the diagram with the given data and solve the problem accordingly.

Complete step-by-step answer:

In the given trapezium ABCD, \[\overline {AB} \parallel \overline {CD} \], \[M\] and \[N\]are the points on the traversals \[\overleftrightarrow {AD}\] and \[\overleftrightarrow {BC}\] respectively as shown in the below diagram.

Also, given \[\dfrac{{AM}}{{MD}} = \dfrac{{BN}}{{NC}} = \dfrac{2}{3}\]

So, clearly from the diagram, \[\overline {MN} \parallel \overline {AB} \] and \[\overline {AB} \parallel \overline {CD} \].

Given that diagonal \[\overline {AC} \] intersects \[\overline {MN} \] at \[O\].

In \[\Delta ADC\], \[\overline {MO} \parallel \overline {DC} \] such that \[M \in \overline {AD} {\text{ }}\& {\text{ }}O \in \overline {AC} \].

We know that by the Triangle Proportionality theorem, if a line parallel to one side of a triangle intersects the other two sides of the triangle, then the lines divide these two sides proportionally.

By using this property in \[\Delta ADC\], we have

\[ \Rightarrow \dfrac{{AM}}{{MD}} = \dfrac{{AO}}{{OC}}\]

But we have \[\dfrac{{AM}}{{MD}} = \dfrac{2}{3}\]

So, \[\dfrac{{AO}}{{OC}} = \dfrac{2}{3}\]

We have to find \[\dfrac{{AO}}{{AC}}\]. From the diagram, \[OC = AO + OC\]

\[

\Rightarrow \dfrac{{AO}}{{OC}} = \dfrac{{AO}}{{AO + OC}} \\

\Rightarrow \dfrac{{AO}}{{OC}} = \dfrac{2}{{2 + 3}} \\

\therefore \dfrac{{AO}}{{OC}} = \dfrac{2}{5} \\

\]

Thus, \[\dfrac{{AO}}{{OC}} = \dfrac{2}{5}\].

Note: The length of the mid-segment is equal to half of the sum of parallel bases in a trapezium. In this problem we have used both the properties of triangles as well as the properties of trapezium.

Complete step-by-step answer:

In the given trapezium ABCD, \[\overline {AB} \parallel \overline {CD} \], \[M\] and \[N\]are the points on the traversals \[\overleftrightarrow {AD}\] and \[\overleftrightarrow {BC}\] respectively as shown in the below diagram.

Also, given \[\dfrac{{AM}}{{MD}} = \dfrac{{BN}}{{NC}} = \dfrac{2}{3}\]

So, clearly from the diagram, \[\overline {MN} \parallel \overline {AB} \] and \[\overline {AB} \parallel \overline {CD} \].

Given that diagonal \[\overline {AC} \] intersects \[\overline {MN} \] at \[O\].

In \[\Delta ADC\], \[\overline {MO} \parallel \overline {DC} \] such that \[M \in \overline {AD} {\text{ }}\& {\text{ }}O \in \overline {AC} \].

We know that by the Triangle Proportionality theorem, if a line parallel to one side of a triangle intersects the other two sides of the triangle, then the lines divide these two sides proportionally.

By using this property in \[\Delta ADC\], we have

\[ \Rightarrow \dfrac{{AM}}{{MD}} = \dfrac{{AO}}{{OC}}\]

But we have \[\dfrac{{AM}}{{MD}} = \dfrac{2}{3}\]

So, \[\dfrac{{AO}}{{OC}} = \dfrac{2}{3}\]

We have to find \[\dfrac{{AO}}{{AC}}\]. From the diagram, \[OC = AO + OC\]

\[

\Rightarrow \dfrac{{AO}}{{OC}} = \dfrac{{AO}}{{AO + OC}} \\

\Rightarrow \dfrac{{AO}}{{OC}} = \dfrac{2}{{2 + 3}} \\

\therefore \dfrac{{AO}}{{OC}} = \dfrac{2}{5} \\

\]

Thus, \[\dfrac{{AO}}{{OC}} = \dfrac{2}{5}\].

Note: The length of the mid-segment is equal to half of the sum of parallel bases in a trapezium. In this problem we have used both the properties of triangles as well as the properties of trapezium.

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