As according to given diagram:
AC= 7cm and AB=25 cm and CB= 24 cm.
SO according to triangle ABC:
[tex]\begin{gathered} \sin (\angle CAB)=\frac{CB}{AB} \\ \sin (\angle CAB)=\frac{24}{25} \\ \sin (\angle CAB)=0.96 \\ (\angle CAB)=\sin ^{-1}(0.96) \\ (\angle CAB)=73.7 \end{gathered}[/tex]Now given that Angle CAD and BAD are equal so:
[tex]\angle CAD=\angle BAD=\frac{\angle CAB}{2}=\frac{73.7}{2}=36.85[/tex]Now in triangle ACD:
[tex]\begin{gathered} \tan (\angle CAD)=\frac{CD}{AC} \\ 7\times\tan (36.85)=CD \\ CD=5.246 \end{gathered}[/tex]And :
[tex]undefined[/tex]