A Question is given followed by two Statements I and II. Consider the Question and the Statements and mark the correct option.Question : ABCD is a quadrilateral such that AC bisects $\angle A$ and $\angle C$. Is AB/DC equal to AD/BC ?Statement-I : AB is parallel to DCStatement-II : AD is parallel to BCWhich one of the following is correct in respect of the above Question and the Statements ?

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CDS - 2026

A Question is given followed by two Statements I and II. Consider the Question and the Statements and mark the correct option.
Question : ABCD is a quadrilateral such that AC bisects $\angle A$ and $\angle C$. Is AB/DC equal to AD/BC ?
Statement-I : AB is parallel to DC
Statement-II : AD is parallel to BC
Which one of the following is correct in respect of the above Question and the Statements ?

This question was previously asked in

CDS 2 2026 Maths Question Paper (13-Sep-2026)

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The Question can be answered by using one of the statements alone, but cannot be answered using the other statement alone

The Question can be answered by using either Statement alone

The Question can be answered by using both the Statements together, but cannot be answered using either Statement alone

The Question can be answered even without using both the Statements

The correct answer is

The Question can be answered even without using both the Statements

Understanding the Condition

The problem involves a quadrilateral ABCD where the diagonal AC bisects both angle A ($\angle DAB$) and angle C ($\angle BCD$). This implies:

$\angle BAC = \angle DAC$
$\angle BCA = \angle DCA$

Let's denote $\angle BAC = \angle DAC = \alpha$ and $\angle BCA = \angle DCA = \beta$.

Applying the Sine Rule

Using the Sine Rule in the triangles formed by the diagonal AC:

In $\triangle ABC$: $\frac{AB}{\sin(\angle BCA)} = \frac{BC}{\sin(\angle BAC)}$ which becomes $\frac{AB}{\sin \beta} = \frac{BC}{\sin \alpha}$.
In $\triangle ADC$: $\frac{AD}{\sin(\angle DCA)} = \frac{DC}{\sin(\angle DAC)}$ which becomes $\frac{AD}{\sin \beta} = \frac{DC}{\sin \alpha}$.

From these, we derive the ratios:

From $\triangle ABC$: $\frac{AB}{BC} = \frac{\sin \beta}{\sin \alpha}$.
From $\triangle ADC$: $\frac{AD}{DC} = \frac{\sin \beta}{\sin \alpha}$.

Therefore, a direct consequence of AC bisecting $\angle A$ and $\angle C$ is that $\frac{AB}{BC} = \frac{AD}{DC}$.

Evaluating the Question

The question asks whether $\frac{AB}{DC} = \frac{AD}{BC}$.

This equality can be rewritten as $AB \cdot BC = AD \cdot DC$.

We established that $\frac{AB}{BC} = \frac{AD}{DC}$. Let this common ratio be $k$. This means $AB = k \cdot BC$ and $AD = k \cdot DC$. Substituting these into the equality $AB \cdot BC = AD \cdot DC$:

Is $(k \cdot BC) \cdot BC = (k \cdot DC) \cdot DC$ ?

This simplifies to $k \cdot BC^2 = k \cdot DC^2$. Assuming $k \neq 0$, this equality holds only if $BC^2 = DC^2$, meaning $BC = DC$.

Conclusion on Sufficiency

The condition $BC = DC$ is not guaranteed solely by the fact that AC bisects $\angle A$ and $\angle C$. Therefore, the statement $\frac{AB}{DC} = \frac{AD}{BC}$ is not always true.

Since we can determine that the statement $\frac{AB}{DC} = \frac{AD}{BC}$ is not necessarily true based only on the given information (AC bisects $\angle A$ and $\angle C$), the question can be answered without needing to refer to Statement I or Statement II.

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