UPSC CDS 1 2022 Math Paper

Concept:In a circle, the perpendicular bisector of a chord passes through the centre; for two concentric circles, a line intersecting them creates symmetrical segments because the perpendicular from the centre bisects both chords.Explanation:For statement 1: Let $AB = AC$ be two equal chords. Draw the angle bisector $AP$ of $\angle CAB$. In $\triangle APB$ and $\triangle APC$: $AB = AC$ (given), $\angle BAP = \angle CAP$ (by construction), and $AP = AP$ (common). So $\triangle APB \cong \triangle APC$ by SAS congruence. Hence $BP = CP$ and $\angle APB = \angle APC$. Since $\angle APB + \angle APC = 180^{\circ}$ (linear pair), $2\angle APB = 180^{\circ}$ or $\angle APB = 90^{\circ}$. Thus $AP$ is the perpendicular bisector of chord $BC$. The perpendicular bisector of a chord passes through the centre; therefore the centre lies on $AP$, the angle bisector. Statement 1 is correct.For statement 2: Let two concentric circles have centre $O$. A line meets the outer circle at $A$ and $D$, and the inner circle at $B$ and $C$. Draw $OM \perp AD$. Then $M$ is the midpoint of chord $AD$ (outer circle) and also midpoint of chord $BC$ (inner circle) because the perpendicular from the centre bisects a chord. So $AM = MD$ and $BM = MC$. Subtracting: $AM - BM = MD - MC$ gives $AB = CD$. Adding $BC$ to both sides: $AB + BC = CD + BC$ gives $AC = BD$. Statement 2 is correct.Answer:Both statements 1 and 2 are correct. Hence option C is the correct answer.