I won't write the detailed process. It's too difficult to tell you the idea.

15. According to SSS, triangle ABC is equal to triangle ADC angle ABC= angle ADC, so angle EDC= angle CBF, BC=CD.

Angle CED= angle CFB = 90°, so all triangles CDE are equal to triangle CBF, so CE = CF

After 16.D, if DE is perpendicular to BC and crosses BC at point E, there will be AD=DE, then triangle ABC, two right angles of triangle EDC, and angle C, so the remaining angle EDC= Angle ABC= Angle C, so ED=EC, that is, AD=ED=EC, and then it will be proved that triangle ABD is all equal to EBD.

So AB=BE and finally BC=BE+EC=AB+AD.