Angle Bisect

[nycmath]

See attachment.

[Eigenvalue]

[tex]\vec{AC}[/tex] bisects [tex]\angle[/tex]BAD, therefore [tex]\angle[/tex]DAC=[tex]\angle[/tex]CAB and [tex]\angle[/tex]DAC+[tex]\angle[/tex]CAB=[tex]\angle[/tex]BAD

m[tex]\angle[/tex]DAC=[tex]\frac{7}{2}[/tex]x°
m[tex]\angle[/tex]CAB=(3x+5)°

Substitute in the values for [tex]\angle[/tex]DAC and [tex]\angle[/tex]CAB
[tex]\frac{7}{2}[/tex]x°=(3x+5)°
[tex]\frac{7}{2}[/tex]x=3x+5
7x=2(3x+5)
7x=6x+10
x=10
m[tex]\angle[/tex]DAC=m[tex]\angle[/tex]CAB=35°

[nycmath]

[tex]\vec{AC}[/tex] bisects [tex]\angle[/tex]BAD, therefore [tex]\angle[/tex]DAC=[tex]\angle[/tex]CAB and [tex]\angle[/tex]DAC+[tex]\angle[/tex]CAB=[tex]\angle[/tex]BAD

m[tex]\angle[/tex]DAC=[tex]\frac{7}{2}[/tex]x°
m[tex]\angle[/tex]CAB=(3x+5)°

Substitute in the values for [tex]\angle[/tex]DAC and [tex]\angle[/tex]CAB
[tex]\frac{7}{2}[/tex]x°=(3x+5)°
[tex]\frac{7}{2}[/tex]x=3x+5
7x=2(3x+5)
7x=6x+10
x=10
m[tex]\angle[/tex]DAC=m[tex]\angle[/tex]CAB=35°

Questions like this take me back to my high school days. I can't believe that I graduated from high school 42 years ago.
It's scary to know that 1984 will never come again. I posted lots of new problems today. Enjoy. See you tomorrow at
7 pm. I hope you had a wonderful weekend.