Math Is Fun Forum

denis_gylaev

Member

Registered: 2015-03-19

Posts: 66

Very Difficult Angle Bisector Problems

Hi;
I am having a very hard time with these 2 problems and I've been stuck on them for 45 minutes now
https://gyazo.com/e89466b148416710e81c54e5ca8cbd4d
https://gyazo.com/4b1403e1f9d5275faee200f4423655ff

apsara123

Member

Registered: 2016-05-14

Posts: 20

Re: Very Difficult Angle Bisector Problems

1)From $\triangle TIU$, we have $\angle ITU + \angle IUT + 109^\circ = 180^\circ$, so
\[\angle ITU  + \angle IUT = 71^\circ.\]
Since $\overline{TX}$ and $\overline{UY}$ are angle bisectors, we have $\angle ITU = \frac{\angle VTU}{2}$ and $\angle IUT = \frac{\angle VUT}{2}$. So, we have
\[\frac{\angle VTU}{2} + \frac{\angle VUT}{2} = 71^\circ,\]
which gives us $\angle VTU + \angle VUT = 142^\circ$. Finally, from $\triangle VTU$, we have
\[\angle V = 180^\circ - \angle VTU - \angle VUT = \boxed{38^\circ}.\]

2)From the Angle Bisector Theorem, we have $BC/BF = AC /AF$. Since $AF = AB - BF = 8-BF$, $AC = AE +CE = 10$, and $BC = 16/3$ (from the previous part), we have $(16/3)/BF = 10/(8-BF)$. Therefore, we have $30BF = 128 - 16BF$, so $46BF = 128$, from which we have $BF = \boxed{64/23}$.

You're having a hard time with these problems...i spent weeks trying to solve these (i was doing these problems before you posted)

denis_gylaev

Member

Registered: 2015-03-19

Posts: 66

Re: Very Difficult Angle Bisector Problems

Thank you for the help; I really understand it now. Just needed a jumping off point.