Question

ABCD is a trapezium whose BC parallel AD and $AD = 4\,cm$. Diagonals AC and BD are intersected at O in such way that $\dfrac{{AO}}{{OC}} = \dfrac{{DO}}{{OB}} = \dfrac{1}{2}$, then what is the length of BC?

Answer 1

Hint: In this question, first we will prove two triangles as similar which have parallel sides as their bases. Then we will use the property of similar triangles of proportional sides to find the length of base BC.

Complete solution:
        

In the above question it is given that AD is parallel to side BC.
Also, $\angle BOC = \angle AOD = x$ (by vertically opposite angle)
And $\angle OBC = \angle ODA = y$ (by alternate angles).
Also, it is given that $\dfrac{{AO}}{{OC}} = \dfrac{{DO}}{{OB}} = \dfrac{1}{2}$.
Therefore, $OB = 2OD$
Let’s take the length of OD be a.
Then the length of $OB = 2a$.
Now, in $\vartriangle ODA\,\,and\,\,\vartriangle OBC$
$\angle BOC = \angle AOD$
$\angle OBC = \angle ODA$
Therefore, $\vartriangle ODA\,\, \sim \,\,\vartriangle OBC$(both the triangles are similar)
Therefore,
$\dfrac{{OB}}{{BC}} = \dfrac{{OD}}{{AD}}$
$ \Rightarrow \dfrac{{2a}}{m} = \dfrac{a}{4}$
Now on simplification we get
$ \Rightarrow m = 8cm$

Therefore, the length of BC is $8cm.$

Note:
In this question we can also prove that triangle OAB is similar to triangle OCD using the same property. There are some properties of a trapezium as follows: In trapezium, exactly one pair of opposite sides are parallel. The diagonals intersect each other. The non-parallel sides in the trapezium are unequal except in isosceles trapezium. The line that joins the mid-points of the non-parallel sides is always parallel to the bases or parallel sides which is equal to half of the sum of parallel sides. In isosceles trapezium, the legs or non-parallel sides are congruent. The sum of the two adjacent angles is equal to \[180^\circ \]. This means that the two adjacent angles are supplementary.