Question

The diagonals of a quadrilateral ABCD intersect each other at the point O such that $\frac{AO}{OC} = \frac{BO}{OD}$. Show that quadrilateral ABCD is a trapezium.

Hint:

Remember that the converse is also true: if $ABCD$ is a trapezium with $AB \parallel CD$, then its diagonals divide each other proportionally ($\frac{AO}{OC} = \frac{BO}{OD}$).
This symmetric relationship is extremely common in geometry problems, and mastering this proof is highly recommended for board exams.

Answer 1

Step 1: Look at the two triangles formed by the diagonals, instead of drawing an extra parallel line.
The diagonals $AC$ and $BD$ intersect at $O$, forming two triangles, $\Delta AOB$ and $\Delta COD$, that face each other across the point $O$.

Step 2: Note the equal vertically opposite angles.
Since $AC$ and $BD$ are straight lines crossing at $O$:
\[ \angle AOB = \angle COD \quad \text{(vertically opposite angles)} \]

Step 3: Rearrange the given ratio to match the SAS similarity condition.
We are given:
\[ \frac{AO}{OC} = \frac{BO}{OD} \]
This can be rewritten by cross-multiplying differently, as:
\[ \frac{AO}{CO} = \frac{BO}{DO} \]
This says that the two sides of $\Delta AOB$ meeting at $O$ (namely $AO$ and $BO$) are proportional to the two corresponding sides of $\Delta COD$ meeting at $O$ (namely $CO$ and $DO$), in the same order.

Step 4: Conclude similarity of the two triangles by SAS.
Since two sides of $\Delta AOB$ are proportional to two sides of $\Delta COD$, and the angle between them ($\angle AOB$ and $\angle COD$) is equal, by the SAS similarity criterion:
\[ \Delta AOB \sim \Delta COD \]

Step 5: Use this similarity to get a pair of equal angles.
Since the triangles are similar, their corresponding angles are equal. In particular, the angle at $A$ in $\Delta AOB$ equals the angle at $C$ in $\Delta COD$:
\[ \angle OAB = \angle OCD \]
That is:
\[ \angle CAB = \angle ACD \]

Step 6: Recognize these as alternate interior angles for the transversal AC.
The diagonal $AC$ acts as a transversal cutting across the lines $AB$ and $CD$. The angles $\angle CAB$ (at vertex $A$, between line $AB$ and transversal $AC$) and $\angle ACD$ (at vertex $C$, between line $CD$ and transversal $AC$) are alternate interior angles with respect to lines $AB$ and $CD$.
Since these alternate interior angles are equal, the two lines $AB$ and $CD$ must be parallel:
\[ AB \parallel CD \]

Final Answer:
Since one pair of opposite sides, $AB$ and $CD$, is parallel, quadrilateral $ABCD$ is by definition a trapezium. Hence Proved.