use the diagram shown. find eg.
eg=\boxed{}
(simplify your answer.)

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Accepted Answer

# Explanation:
## Step1: Recognize the Angle Bisector Theorem (or related property)
We know that in a triangle, if a point is equidistant from the sides of an angle (angle bisector property) and also, the segments from the point to the sides (perpendiculars) relate to the sides of the triangle. Here, we can use the property of similar triangles or the ratio of segments. Notice that $ DG = 11 $, $ BG = 14 $, and $ BC = 42 $. Let's consider the ratio of $ BE $ to $ BC $ and the ratio of $ DG $ to $ EG $ (or vice - versa, depending on the similar triangles). Wait, actually, since $ G $ is a point such that $ GD\perp AB $, $ GE\perp BC $, and $ GF\perp AC $, and $ BA $ and $ BC $ are sides with $ BG $ as a common segment. Also, we can use the concept of the ratio of the lengths of the perpendiculars and the sides.

Let's assume that triangles $ BDG $ and $ BEG $ are related? Wait, no. Wait, the key is that the ratio of $ DG $ to $ EG $ should be equal to the ratio of $ AB $ to $ BC $? No, maybe better to use the area or the angle bisector with perpendiculars.

Wait, another approach: Since $ G $ is on the angle bisector of $ \angle B $ (because $ GD\perp AB $ and $ GE\perp BC $, so by the angle bisector theorem's converse, $ GD = $ distance from $ G $ to $ AB $, $ GE=$ distance from $ G $ to $ BC $, but actually, the ratio of $ DG $ to $ EG $ is equal to the ratio of $ AB $ to $ BC $? No, maybe the triangles $ BDG $ and $ BEG $ are similar? Wait, $ \angle BDG=\angle BEG = 90^{\circ} $, and $ \angle DBG=\angle EBG $ (since $ BG $ is the angle bisector). So triangles $ BDG $ and $ BEG $ are similar by AA similarity.

So, $ \frac{DG}{EG}=\frac{BD}{BE} $? Wait, no, in similar triangles, corresponding sides are proportional. So $ \frac{DG}{EG}=\frac{BD}{BE} $, but we don't know $ BD $ and $ BE $. Wait, maybe we can use the fact that the ratio of the lengths of the perpendiculars from $ G $ to the sides $ AB $ and $ BC $ is related to the ratio of the lengths of the segments on $ BC $ and $ AB $. Wait, alternatively, notice that the length of $ BC = 42 $, and let's assume that the ratio of $ DG $ to $ EG $ is equal to the ratio of $ AB $ to $ BC $? No, maybe the key is that $ \frac{DG}{EG}=\frac{11}{EG}=\frac{BD}{BE} $, but we know that $ BG $ is common. Wait, maybe the triangles $ BDG $ and $ BEG $ are similar, so $ \frac{DG}{EG}=\frac{BD}{BE} $, but we can also use the fact that the ratio of $ DG $ to $ EG $ is equal to the ratio of the adjacent sides. Wait, another way: Let's use the formula for the length of the perpendicular from a point on the angle bisector.

Wait, actually, the correct approach is: Since $ GD\perp AB $ and $ GE\perp BC $, and $ BG $ is the angle bisector of $ \angle B $, then $ \frac{DG}{EG}=\frac{AB}{BC} $? No, that's not right. Wait, no, the angle bisector theorem for perpendiculars: the distance from a point on the angle bisector to the two sides is equal? No, that's only if the point is on the angle bisector and the two sides are equal, but here $ AB $ and $ BC $ are not necessarily equal. Wait, no, the distance from a point on the angle bisector to the two sides is equal. Wait, that's the angle bisector theorem: any point on the angle bisector is equidistant from the two sides. Wait, that's a key property! So $ GD = GE $? No, that can't be, because $ DG = 11 $ and if that's the case, $ EG = 11 $, but that seems wrong. Wait, no, wait: the angle bisector of $ \angle B $ - the distance from $ G $ to $ AB $ (which is $ DG $) and the distance from $ G $ to $ BC $ (which is $ EG $) should be equal. But that would mean $ DG = EG $, so $ EG = 11 $? But that contradicts the length of $ BC = 42 $. Wait, maybe I made a mistake. Wait, no, the angle bisector theorem says that a point on the angle bisector is equidistant from the two sides. So if $ BG $ is the angle bisector of $ \angle B $, then $ GD = GE $, because $ GD\perp AB $ and $ GE\perp BC $. So $ EG=DG = 11 $? But that seems too simple. Wait, but let's check the diagram again. The diagram shows $ DG = 11 $, $ BG = 14 $, $ BC = 42 $. Wait, maybe the angle is not $ \angle B $ but another angle. Wait, no, $ GD $ is perpendicular to $ AB $, $ GE $ is perpendicular to $ BC $, so $ G $ is on the angle bisector of $ \angle B $, so $ GD = GE $. Therefore, $ EG = 11 $. Wait, but that seems too easy. Alternatively, maybe we use the geometric mean or something else. Wait, another approach: Let's consider the area of triangle $ BCG $ and triangle $ BDG $. Wait, no, maybe the triangles $ BDG $ and $ BEG $ are similar, and we can use the ratio of $ BG $ and the sides. Wait, $ BG $ is common, $ \angle BDG=\angle BEG = 90^{\circ} $, $ \angle DBG=\angle EBG $, so triangles $ BDG \sim BEG $ (AA). Therefore, $ \frac{DG}{EG}=\frac{BD}{BE} $, but we don't know $ BD $ and $ BE $. Wait, but maybe the ratio of $ DG $ to $ EG $ is equal to the ratio of $ AB $ to $ BC $? No, that doesn't make sense. Wait, maybe the length of $ BC = 42 $, and $ DG = 11 $, and we can use the proportion $ \frac{DG}{EG}=\frac{11}{EG}=\frac{BD}{BE} $, but we don't have $ BD $ and $ BE $. Wait, maybe the key is that the distance from $ G $ to $ AB $ (DG) and to $ BC $ (EG) are related by the ratio of the lengths of $ AB $ and $ BC $, but since $ BG $ is the angle bisector, the ratio of $ DG $ to $ EG $ is 1 (since it's the angle bisector, the distances are equal). Therefore, $ EG = DG=11 $.

## Step2: Confirm the property
The angle bisector theorem states that any point on the angle bisector of an angle is equidistant from the two sides of the angle. Since $ G $ lies on the angle bisector of $ \angle B $ (as $ GD\perp AB $ and $ GE\perp BC $), the distance from $ G $ to $ AB $ (which is $ DG = 11 $) is equal to the distance from $ G $ to $ BC $ (which is $ EG $). Therefore, $ EG=11 $.

# Answer:
$ 11 $

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Explanation:

Step1: Recognize the Angle Bisector Theorem (or related property)

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