Question

ABCD is a trapezium with $AB\parallel DC$ , and its diagonals intersect each other at O. Show that $\dfrac{AO}{BO}=\dfrac{CO}{DO}$

Answer 1

Hint: Draw a line parallel to AB through O intersecting BD at E. Apply basic proportionality theorem in triangles ABC and BCD. Hence find the relations for $\dfrac{AO}{OC}$ and $\dfrac{BO}{OD}$. Compare the two relations and hence prove that $\dfrac{AO}{BO}=\dfrac{CO}{DO}$.

Complete step-by-step answer:

Given: ABCD is a trapezium with $AB\parallel CD$. Diagonals AC and BD intersect each other at O.
To prove: $\dfrac{AO}{BO}=\dfrac{CO}{DO}$
Construction: Draw a line parallel to AB through O and let it intersect BC at E.
Proof:
We know that $AB\parallel OE$ and $AB\parallel CD$.
Hence $OE\parallel CD$.
Now in triangle ABC, we have $OE\parallel AB$(By construction).
We know that a line parallel to one side of a triangle divides the other two sides proportionately (Basic proportionality theorem).
Hence by basic proportionality theorem, we have
$\dfrac{AO}{CO}=\dfrac{BE}{CE}\text{ (i)}$
Also, in triangle BCD, we have $OE\parallel CD$
Hence by basic proportionality theorem, we have
$\dfrac{BO}{DO}=\dfrac{BE}{CE}\text{ (ii)}$
Hence from equation (i) and equation (ii), we get
$\dfrac{AO}{CO}=\dfrac{BO}{DO}$
Multiplying both sides of the equation by $\dfrac{CO}{BO}$, we get
$\begin{align}
  & \dfrac{AO}{CO}\times \dfrac{CO}{BO}=\dfrac{BO}{DO}\times \dfrac{CO}{BO} \\
 & \Rightarrow \dfrac{AO}{BO}=\dfrac{CO}{DO} \\
\end{align}$
Hence proved.

Note: Alternative method:
Since $AB\parallel CD$, we have
$\angle DCO=\angle OAB$ (alternate interior angles) and $\angle ODC=\angle OBA$ (alternate interior angles).
Hence In triangles OCD and OAB, we have
$\angle DCO=\angle OAB$ (proved above)
$\angle ODC=\angle OBA$(proved above)
$\angle COD=\angle AOB$ (Vertically opposite angles).
Hence, $\Delta COD\sim \Delta AOB$ by AAA similarity criterion.
Hence $\dfrac{AO}{CO}=\dfrac{BO}{DO}$
Multiplying both sides by $\dfrac{CO}{BO}$, we get
$\begin{align}
  & \dfrac{AO}{CO}\times \dfrac{CO}{BO}=\dfrac{BO}{DO}\times \dfrac{CO}{BO} \\
 & \Rightarrow \dfrac{AO}{BO}=\dfrac{CO}{DO} \\
\end{align}$
Hence proved.