A parallelogram cant be a cross section of a triangular prism? never true,sometimes true,or always true

Mathematics

1 answer:

Sedaia [141]3 years ago

8 0

Answer:

I wanna say never true...Hope thats right

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Pleaseeeeeeeeeee help meeeeee

Aloiza [94]

The answer is c thank me later

4 0

2 years ago

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

vova2212 [387]

Answer:

▢ $\overline{DG}$ ≅ $\overline{FG}$

▢ $\overline{DG}$ ≅ $\overline{BG}$

▢ $\overline{GE}$ bisects $\displaystyle ∠DEF$

▢ $\overline{GA}$ bisects $\displaystyle ∠BAF$

Step-by-step explanation:

As you can see, angles $\displaystyle BAF$ and $\displaystyle DEF$ form two half-kites, which you should know have two pairs of congruent sides, therefore segments $\displaystyle GA$ and $\displaystyle GE$ are what you call angle bisectours, which is why segments $\displaystyle FG,$ $\displaystyle DG,$ and $\displaystyle BG$ are congruent.