Question

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

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 $△ADC$,
Since $EO\parallel CD$ ( from construction )
$\Rightarrow {\displaystyle \frac{AE}{ED}}={\displaystyle \frac{AO}{OC}}$ ( By Basic Proportionality Theorem ) (i)
Also in $△ADB$
$\Rightarrow {\displaystyle \frac{AE}{ED}}={\displaystyle \frac{BO}{OD}}$ ( By Basic Proportionality Theorem ) (ii)
Now comparing equations (i) and (ii)
${\displaystyle \frac{AO}{OC}}={\displaystyle \frac{BO}{OD}}$
$\Rightarrow {\displaystyle \frac{AO}{BO}}={\displaystyle \frac{CO}{OD}}$ ( 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.