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Given - A triangle ABC in which AB = 26 , BC = 28 , CA = 30 . AD is the internal angle bisector of angle A . BE is perpendicular from B to AD . G is a point on BC such that EG || CA . To Find :- DG .

Construction :- Draw AF perpendicular to BC .

AD is the angle bisector of angle A . .'. BD:DC = AB : AC { By angle bisector theorem }

=> BD : DC = 26 : 30 = 13 : 15 And, also BD + DC = BC = 28
So, BD = $\frac{13}{28}$ * 28 = 13
.'. DC = BC - BD =28 - 13 = 15

In the adjoining figure , Ar. triangle ABC = 336 { Calculate it using Heron's formula } => 1/2 * BC * AF = 336 => 1/2 * 28 * AF = 336 => AF = 24

In right angled triangle AFB , BF^2 = AB^2 - AF^2 {by Pythagoras' theorem}
=> BF = √(26^2 - 24^2) = √(100) = 10
Now , DF = BD - BF = 13 - 10 = 3

In right triangle AFD , AD^2 = AF^2 + DF ^2 {by Pythagoras' theorem} AD = √(24^2 + 3^2) = √(585)

.'. Ar. triangle ABD = $\frac{1}{2}$ * AF * BD = $\frac{1}{2}$ * 24 * 13 = 156

=> 1/2 * AD*BE = 156 { '.' AREA OF TRIANGLE ABD = 1/2 *AD *BE }

=> 1/2 * √585 * BE = 156 => BE = $\frac{312}{√585}$

In right triangle BED, ED^2 = BD^2 - BE^2 { by Pythagoras' theorem}
ED = √(13^2 - $\frac{312}{√585}$ ^2) = 1.6124515497

Now in triangle ACD , GE || CA .'. $\frac{ED}{AD}$ = $\frac{DG}{CD}$ { by sir Thales' theorem} => $\frac{1.6124515497}{√585}$ = $\frac{DG}{15}$ => DG = 1

.................................................................................................................................................................. by Vishwash But the one which is below (by sir ahmad saad) is the best .

$CosA=\dfrac{c^2+b^2 - a^2}{2*c*b}.~~~~\\ AE=c*Cos(\frac 1 2*Cos^{-1}A)=22.5743.\\ DC=\dfrac {28}{26+30}*30=15.\\ \therefore~BD=28 - 15=13.\\ CosB=\dfrac{26^2+28^2 - 30^2}{2*26*28}.\\ AD=\sqrt{AB^2+BD^2- 2*AB*BD*CosB}\\ =\sqrt{26^2+13^2 - 2*26*13*\dfrac{26^2+28^2 - 30^2}{2*26*28}} =24.18677.\\ \Delta~ADC{ \Large \text{ ~} } ~\Delta~EDG.~~~\\ \implies~DG=\dfrac{DC*DE}{AD}\\ = \dfrac{15*(AD-AE)}{AD}=1.0000013455$