Question

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

Answer 1

Hint: Draw a line parallel to AB and DC . Using the Basic Proportionality Theorem and the constructed triangles inside the trapezium prove the required answer.

Complete step-by-step answer:

In trapezium ABCD with $AB\parallel DC$, drawing a line $EF\parallel CD$
Now according to Basic Proportionality Theorem which states that "If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio".
Now in $\vartriangle ADC$,
Since $EO\parallel CD$ ( from construction )
$ \Rightarrow \dfrac{{AE}}{{ED}} = \dfrac{{AO}}{{OC}}$ ( By Basic Proportionality Theorem ) (i)
Also in $\vartriangle ADB$
$ \Rightarrow \dfrac{{AE}}{{ED}} = \dfrac{{BO}}{{OD}}$ ( By Basic Proportionality Theorem ) (ii)
Now comparing equations (i) and (ii)
$\dfrac{{AO}}{{OC}} = \dfrac{{BO}}{{OD}}$
$ \Rightarrow \dfrac{{AO}}{{BO}} = \dfrac{{CO}}{{OD}}$ ( cross multiplying )
Hence proved.

Note: Recall Basic Proportionality Theorem to solve such types of questions. Construction becomes important in solving such questions in a simple manner. We should make constructions wherever required.