Posted by **luis** on Thursday, September 19, 2013 at 2:21am.

Given: FG (SEG) bisects BT (SEG)

BO=7x-6

OT=5x+10

Prove: BO=50

- geometry -
**Graham**, Thursday, September 19, 2013 at 3:08am
There appears to be insufficient information. Where is O, with respect to BT and FG?

- geometry -
**Steve**, Thursday, September 19, 2013 at 4:06am
It appears to me that O is midway between B and T. If so, then

7x-6 = 5x+10

2x = 16

x = 8

So, BO = 5*8+10 = 50

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