Question

In a trapezium $ABCD$, $AB$ is parallel to $DC$ and the diagonals intersect at $O$. Show that $OA:OC=OB:OD$. If $AB=2DC$, show that $O$ is the point of trisection of both the diagonals.

Answer 1

Step 1: Construct a trapezium using the given data:

Given that a trapezium $ABCD$, $AB$ is parallel to $DC$ and the diagonals intersect at $O$.

The trapezium can be drawn as,

Draw a line $EF$through point $O$, such that$EF$is parallel to $AB$ and $DC$.

Step 2: Use the basic proportionality theorem on triangle $ADC$ and triangle $AEO$:

Basic proportionality Theorem:

If in a triangle, a line drawn parallel to one side, intersects the other side in distinct points divides two sides in the same ratio.

Consider $△ADC$ and $△AEO$.

$OE$ is parallel to $DC$.

According to the basic proportionality theorem,

$\frac{AE}{ED}=\frac{OA}{OC}$ …..(i)

Step 3: Use the basic proportionality theorem on triangle $ABD$ and $AEO$:

Basic proportionality Theorem:

If in a triangle, a line drawn parallel to one side, to intersect the other side in distinct points divides the two sides in the same ratio.

Consider, $△ABD$ and $△DEO$.

$OE$ is parallel to $AB$.

According to the basic proportionality theorem,

$\frac{AE}{ED}=\frac{OB}{OD}$ …..(ii)

From equations (i) and (ii),

$\frac{OA}{OC}=\frac{OB}{OD}$

$OA:OC=OB:OD$

Hence proved.

Step 4: Check whether $∆AOB$ and $∆DOC$ are similar

AAA similarity: If in two triangles, the corresponding angles are equal, then the triangles are similar.

$\angle AOB=\angle DOC\left(\mathrm{Vertically}\mathrm{opposite}\right)$

$\angle ABO=\angle ODC\left(\mathrm{Alternate}\mathrm{interior}\mathrm{angle}\right)$

$\angle OAB=\angle OCD\left(\mathrm{Alternate}\mathrm{interior}\mathrm{angle}\right)$

According to AAA similarity, we get

$∆AOB~$$∆DOC$

Step 5: Prove O is the point of trisection of the diagonals

Similar triangle property: If two triangles are similar then their corresponding sides are proportional.

Given that $AB=2DC$

$∆AOB$ and $∆DOC$ are similar.

So, their sides are in proportion.

$\frac{OC}{OA}=\frac{DC}{AB}=\frac{OD}{OB}$

$\Rightarrow \frac{OC}{OA}=\frac{DC}{2DC}=\frac{OD}{OB}\left[\because AB=2DC\right]$

$\Rightarrow \frac{OC}{OA}=\frac{1}{2}=\frac{OD}{OB}$

$\Rightarrow OA=2OC\mathrm{and}OB=2OD$

Hence $O$ is the point of trisection of the diagonals $AC$ and $BD$.