06 ce maths question!!!

(a)(i)
∠BAS = ∠BCS = 90° (∠ in semi-circle)
Produce CH to meet AB to K
∠BKC = ∠BOA = 90° (orthocentre of Δ)
Hence, ∠BAS = 90°= ∠BKC
and ∠BCS = 90°= ∠BOA
So, AS//HC and SC//AH (corr. ∠s equal)
Thus, AHCS is a parallelogram. (By definition)

(ii)
AHCS is a parallelogram. (by (a)(i) )
∴AH = SC (opp. sides, //gram)
GR⊥BC
∴BR = RC (line from centre⊥chord bisects chord)
BG = GS (radius)
∴SC = 2GR (mid-point theorem)
∴AH = 2GR