In the figure, AD = CD and AB = CB.

AD = CD

AB  = CB

And DB  = DB

So...by  SSS, triangle DAB  is congruent to triangle DCB

But  this implies that  angle BDA  =  angle BDC

So...   BD bisects angle ADC

To prove the second part.....

Since  AD  = CD, then the angles opposite those sides in triangle ADC are also equal

So  angle DAC =  angle DCA

So

angle BDA  = angle BDC

angle DAC  =  angle DCA 

And  DE  = DE

So....by AAS, trianlgle   ADE  is congruent to triangle CDE

But this implies that  angle AED  =  angle CED

But...  m  AED  +  m CED  =  180

So, by substitution   m AED + m AED  = 180

2m AED  =  180       divide by 2 on both sides

m AED  = 90   =  m CED

And a line standing upon a line forming two right angles at their intersection means that the lines are perpendicular....so, .AC is perpendicular to DB