Prove the concyclic(very hard)

a)

∠NAP =∠NA1P = 90˚ (given)
So AA1PN are concyclic (converse of ∠s in the same segment)

b)

M is the mid-pt of BC and B1C1 (given)
△ABC≅△A1B1C1 (given)
BC=B1C1 (corr.sides,≅△s)
BM+MC=B1M+MC1
2MC = 2MC1
MC=MC1
AC=A1C1 (corr.sides,≅△s)
∠C=∠C1 (common ∠)
So △AMC≅△A1MC1 (S.A.S.)
∠MAC=MA1C1 (corr.∠s,≅△s)
∠MAP=MA1P
So AA1PM are concyclic (converse of ∠s in the same segment)

c)

∠APM=∠AA1M (∠s in the same segment)
∠APN=∠AA1N (∠s in the same segment)
∠APM-∠APN=∠AA1M-∠AA1N
∠NPM=∠NA1M
So A1PMN are concyclic (converse of ∠s in the same segment)
∠NMP + ∠NA1P =180˚ (opp. ∠s, cyclic quad.)
∠NMP + ∠NAP= 180˚
So APMN are concyclic (opp.∠s, supp.)

Understand?