In triangles, #\Delta ACE# & #\Delta BDE#, we have
#\angle ACE=\angle BDE \ \quad ( \text{aternate angles})#
# \angle CAE=\angle DBE \ \quad ( \text{aternate angles})#
# \angle AEC=\angle BED \ \quad ( \text{vertically opposite angles})#
Now, from A-A-A similarity, #\Delta ACE # & # \Delta BDE#
are similar triangles
then the ratios of corresponding sides in the similar triangles must be same as follows
#\frac{BE}{AE}=\frac{BD}{AC}#
#\frac{BE}{AE}=\frac{3}{4}#
#\frac{BE}{AE}+1=\frac{3}{4}+1#
#\frac{BE+AE}{AE}=\frac{3+4}{4}#
#\frac{AB}{AE}=\frac{7}{4}#
#\frac{10.5}{AE}=\frac{7}{4}#
#AE=\frac{4\cdot 10.5}{7}#
#AE=6\ cm#
