# geometry

posted by .

Given: FG (SEG) bisects BT (SEG)
BO=7x-6
OT=5x+10

Prove: BO=50

• geometry -

I think we need to see the figure, but looking at the given, I think the point FG bisects and intersects BT at point O. Therefore we can say the lengths of segments BO and OT are equal, or
7x - 6 = 5x + 10
7x - 5x = 10 + 6
2x = 16
x = 8

Susbtituting,
BO = 7x - 6
BO = 7(8) - 6
BO = 56 - 6
BO = 50

Hope this helps~ :3

## Respond to this Question

 First Name School Subject Your Answer

## Similar Questions

1. ### geometry

We are learning about proofs. We drew lines between noncollinear points. Our table showed 2 points >> 1 line segment, 3 point >> 3 line segments, 4 points >> 6 line segments, 5 points >> 10 line seg, 6 points …
2. ### Geometry

Given: ray AD bisects <BAC <DAC=3x, and <BAD=5x-24 Prove: x=12 its a proof and i need help
3. ### Geometry Please Help

Given: QS bisects ∠RQT; ∠R ≅ ∠ T Prove: SQ bisects ∠RST
4. ### Geometry Please Help

Given: QS bisects ∠RQT; ∠R ≅ ∠ T Prove: SQ bisects ∠RST /Users/szeringue/Documents/triangle.doc
5. ### GEOMETRY

GIVEN: TRIANGLE ABC,AD bisects <BAC,and AE=ED, PROVE:AE/AC=BD/BC
6. ### Geometry

Given: angle BAC is congruent to angle ACD, segment BD bisects segment AC, and segment AC bisects angle BCD. Prove: quadrilateral ABCD is a rhombus.
7. ### Math (Geometry)

Given: ABCD is arhombus. prove:AC bisects angle BAD and angel BCD.
8. ### geometry

Given: FG (SEG) bisects BT (SEG) BO=7x-6 OT=5x+10 Prove: BO=50
9. ### geometry

Given: ABC,/ AD bisects <BAC, and /AE=/ED Prove: AE/AC = BD/BC
10. ### Math

We are learning about proofs. We drew lines between noncollinear points. Our table showed 2 points >> 1 line segment, 3 point >> 3 line segments, 4 points >> 6 line segments, 5 points >> 10 line seg, 6 points …

More Similar Questions