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: Understanding the Question:
This question is a proof-based problem from the "Triangles" chapter.
We are given a quadrilateral $ABCD$ whose diagonals $AC$ and $BD$ intersect at point $O$.
The diagonals divide each other in such a way that the ratio of segment lengths satisfies:
\[ \frac{AO}{OC} = \frac{BO}{OD} \] We need to prove that quadrilateral $ABCD$ is a trapezium (which requires proving that at least one pair of opposite sides is parallel, i.e., $AB \parallel CD$).

Step 2: Key Formula or Approach:
We will use the

Basic Proportionality Theorem (Thales' Theorem) and its converse:
1. Draw an auxiliary line parallel to one of the sides through the intersection point $O$.
2. Apply Thales' Theorem to establish side ratios.
3. Use the given diagonal ratio to show that the auxiliary line is also parallel to the opposite side using the Converse of Thales' Theorem.

Step 3: Detailed Explanation:

• Construction:
Draw a line segment $EO$ through point $O$ parallel to side $AB$, meeting side $AD$ at point $E$ ($EO \parallel AB$).

• Applying Basic Proportionality Theorem in $\Delta DAB$:
In triangle $DAB$, we have $EO \parallel AB$ (by construction).
According to the Basic Proportionality Theorem:
\[ \frac{DE}{EA} = \frac{DO}{OB} \] Take the reciprocal of both sides:
\[ \frac{AE}{ED} = \frac{BO}{OD} \quad \text{--- (Equation 1)} \]

• Using the Given Condition:
We are given the relation:
\[ \frac{AO}{OC} = \frac{BO}{OD} \quad \text{--- (Equation 2)} \]

• Comparing Equation 1 and Equation 2:
Since the right-hand sides of both equations are equal, we can equate their left-hand sides:
\[ \frac{AE}{ED} = \frac{AO}{OC} \]

• Applying Converse of Basic Proportionality Theorem in $\Delta ACD$:
In triangle $ACD$, the line segment $EO$ divides the sides $AD$ and $AC$ in the same ratio:
\[ \frac{AE}{ED} = \frac{AO}{OC} \] By the Converse of Thales' Theorem, this means the line $EO$ must be parallel to the base $CD$:
\[ EO \parallel CD \]

• Concluding the Parallelism of Sides:
From our construction, we have $EO \parallel AB$.
From our proof, we have $EO \parallel CD$.
Since both $AB$ and $CD$ are parallel to the same line $EO$, they must be parallel to each other:
\[ AB \parallel CD \]

• Since quadrilateral $ABCD$ has one pair of opposite sides parallel ($AB \parallel CD$), it is by definition a trapezium.


Step 4: Final Answer:
Hence, quadrilateral $ABCD$ is a trapezium.