UPSC CDS 2 2022 Math Paper

Concept:Use properties of a parallelogram, a cyclic quadrilateral, and a linear pair of angles to prove $AE = AD$, then check the second statement.Explanation:In parallelogram $ABCD$, opposite angles are equal: $\angle ABC = \angle ADC = \theta$.Points $A$, $B$, $C$ lie on the circle, so quadrilateral $ABCE$ is cyclic.In a cyclic quadrilateral, the sum of a pair of opposite angles is $180^{\circ}$: $\angle ABC + \angle AED = 180^{\circ}$.Thus $\angle AED = 180^{\circ} - \theta$. …(1)Points $A$, $D$, $E$ are collinear on line $CD$ produced, so $\angle ADE$ and $\angle ADC$ form a linear pair: $\angle ADE + \angle ADC = 180^{\circ}$.Hence $\angle ADE = 180^{\circ} - \theta$. …(2)From (1) and (2), $\angle AED = \angle ADE$.In triangle $ADE$, sides opposite equal angles are equal, so $AE = AD$.Therefore statement 1 is correct.For statement 2: $CD = DE$ would mean $D$ is the midpoint of $CE$ and also the centre of the circle if $CD$ and $DE$ were radii.But $D$ is a vertex of the parallelogram; the circle passes through $A$, $B$, $C$ and not necessarily through $D$.There is no general condition forcing $CD = DE$; it would require a specific configuration that does not hold for all parallelograms.Thus statement 2 is not always true, so it is incorrect.Answer:Only statement 1 is correct. Hence the correct option is A (Only 1).