UPSC 2015 CDS I Math Paper

In ΔAPD and ΔBQC,
AD = BC (Given)
DP = CQ (Distance between parallel sides)
= ∠APD ∠BQC (90°)
∴ ΔAPD ≅ ΔBQC (RHS congruence rule)
⇒ ∠A = ∠B (CPCT)
∠A + ∠D = 180° (Sum of adjacent interior angles is supplementary)
⇒ ∠B + ∠D = 180°
Therefore, trapezium ABCD is cyclic.
Hence, statement (1) is correct.In ΔOAB, OA = OB = AB = r
ΔOAB is an equilateral triangle.
⇒ ∠AOB = 60°
Now,∠AOB = 2∠ACB (Degreemeasure theorem)
∴ 2∠ACB=60°
⇒ ∠ACB = 30°