Triangle ABC is a right triangle with right angle at A.

Since  AY  bisects the 90° angle...     m∠BAY  =  90° / 2  =  45°

So...     m∠BAX  =  45° - 13°  =  32°

And since there are 180° in triangle BAX...   m∠ABX  =  180° - 32° - 90°  =  58°

And since there are 180° in triangle ABC...   m∠BCA  =  180° - 58° - 90°  =  32°

Now..to show why triangle ACZ is isocelese....

Draw a line from  Z  to side  BA  that is parallel to AC.

Draw a line from  Z  to side  AC that is parallel to BA.

Since  Z is the midpoint of  BC,  BZ = ZC .

So we can be sure that triangle BJZ is congruent to triangle ZHC from the AAS rule.

And...  AH  =  JZ  =  HC

So.. we can be sure that triangle AHZ is congruent to triangle CHZ by the SAS rule.

Therefore...   m∠ACZ  =  m∠ZAC =  32°

Excellent thinking, hectictar....!!!!!....this one was a little tricky, for sure....!!!!!

Here's one more thought......dropping a perpendicular from Z to H on  AC means that  ZH is parallel to BA....and whenevever   a segment is drawn parallel to a base, it splits the sides of the  triangle into  equal ratios....that is.....CZ / BZ  =  CH / AH.....but since BZ  = CZ  [ because Z is the midpoint of BC ], then  AH  = CH

Then by SAS, triangle CHZ   is congruent to triangle AHZ.....and since you found that BCA  = 32°  = ZCH  =  ZAH  = ZAC

Obviously.....dropping that perpendicular as you did was the key to the whole thing.....not bad for a 'Bama fan....LOL!!!!

LOL!!!!!!......we'll see......don't get too cocky....!!!!