1. As an auxiliary line, AB intersects AC at point E through the parallel line of point O. Because there is ∠ d = ∠ Abd = 90, there are AB∨CD∨EO, ∠EC=EA=∠CAB, and because O is the midpoint of bd.

Because ∠ b = 90, ∠ OAB+∠ AOB = 90,

Because ∠CAO=∠OAB, ∠ Cao+∠ AOB = 90.

Because ∠ EOB = 90, ∠EOA=∠CAO, EA=EO, and because EC=EA, EO=EC.

So ∠ECO=∠EOC

Because CD∥EO, ∠DCO=∠EOC, ∠ECO=∠DCO,

So OC divides ∠ACD equally

2. As an auxiliary line, make a straight line perpendicular to AC through point O and intersect AC at point F. ..

△ ABO △ AFO is derived from the corner edge (∠ AFO = ∠ ABO = 90, ∠FAO=∠BAO, AO=AO).

Similarly, △ foc △ doc can be derived.

Because CD∥EO, ∠ ACD = ∠ AEO =180-∠ CEO =180-∠ cab = (90-∠ bao)+(90-∞).

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