UPSC 2014 CDS II Math Paper

OC=OA=

AC

2

(Diagonals of the parallelogram bisect each other)
In ΔBCD,P is the mid-point of CD and R is the mid-point of BC.
∴PR||BD (Mid-point theorem)
In Δ CQP and Δ COD
∠CQP = ∠COD  (Corresponding angles)
∠CPQ = ∠CDO  (Corresponding angles)
∴ Δ CQP ~ ΔCOD (AA Similarity)
⇒

CQ

OC

=

CP

CD