Let CD=DE=EA=x
Applying Apollonius theorem in △ABD
BD2+AB2=2{BE2+x2}⇒16+AB2=2(9+x2)⇒AB2=2+2x2 .....(i)
Applying Apollonius theorem in △BCE, we have
BE2+BC2=2{BD2+x2}⇒9+BC2=2(16+x2)⇒BC2=23+2x2 .....(ii)
In right angled △ABC
AB2+BC2=AC2AB2+BC2=(3x)2 .....(iii)
Substituting values of equations (i) and (ii) in equation (iii), we get
2+2x2+23+2x2=9x2⇒25=5x2⇒x=√5$
Thus AC=3x=3√5
We have a√b=3√5
⇒a=3,b=5⇒a+b=8
Hence, the answer is 8.