Question

G is the incenter, or point of concurrency, of the angle bisectors of ΔACE.

Triangle A C E has point G as its incenter. Lines are drawn from the point of the triangle to point G. Lines are drawn from point G to the sides of the triangle to form right angles. Line segments G B, G D, and G F are formed.
Which statements must be true regarding the diagram?

BG ≅ AG
DG ≅ FG
DG ≅ BG
GE bisects ∠DEF
GA bisects ∠BAF

Answer 1

Assume B, D, F are the feet of the perpendiculars from the incenter G to sides AC, CE, AE respectively (so GB ⟂ AC, GD ⟂ CE, GF ⟂ AE).

- BG ≅ AG — False. GB is the inradius (distance from G to side AC); AG is the distance from vertex A to the incenter. They are not generally equal.
- DG ≅ FG — True. GD and GF are both perpendicular distances from G to sides CE and AE, and the incenter is equidistant from all sides (inradius).
- DG ≅ BG — True. Same reason: GD = GB = GF = inradius.
- GE bisects ∠DEF — True. GE is the angle bisector of ∠CEA (G is the incenter). Points D and F lie on CE and AE, so ∠DEF = ∠CEA, which GE bisects.
- GA bisects ∠BAF — True. GA bisects ∠CAE; B and F lie on AC and AE, so ∠BAF = ∠CAE, which GA bisects.