Ex.6.2 Q9 Triangles Solution - NCERT Maths Class 10

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Question

\(ABCD\) is a trapezium in which \(AB || DC\) and its diagonals intersect each other at the point \(O.\) Show that \(\begin{align}\frac{AO}{BO}=\frac{CO}{DO} \end{align}\)

 

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 trapezium \(ABCD\)

\( AB||CD\) ,\(AC\) and  \(BD\) intersect at \(‘O’\)

Construct \(XY\) parallel to \(AB \) and \(CD\) \((XY\parallel AB,(XY \parallel CD)\)through \(‘O’\)

In \(\begin{align}\Delta ABC\end{align}\)

\(OY||AB\) \((\because construction)\)

According to theorem \(6.1\) (BPT)

\[\begin{align}\frac{BY}{CY}=\frac{OA}{OC}\,\,\,.............\rm{(I)}\end{align}\]

In \(\Delta BCD\)

\(\rm OY||CD\; (\because\; \rm{construction})\)

According to theorem \(6.1\) (\(BPT\))

\[\begin{align}\frac{BY}{CY}=\frac{OB}{OD}.................\left( \text{II} \right) \\ \end{align}\]

From \(\rm (I)\) and \(\rm (II)\)

\[\begin{align} \,\,\,\,\,\,\frac{OA}{OC}&=\frac{OB}{OD}\,\,\,\\ \Rightarrow\quad \frac{OA}{OB}&=\frac{OC}{OD} \\ \end{align}\]

 

  
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