Ray BD bisects angle ABC in triangle ABC such that point D lies on segment AC. Segment DE is drawn parallel to segment AB so that point E is between points B and C. Point F is located on ray BC such that point C is between B and F. It is given that BDE=(4x-7) , ACF= (17x+9) and ADB= (11x-4).
(1) Since AB||DE, `/_ABD cong /_BDE` (Alternate Interior Angles).
(2) `m/_B=8x-14` (Definition of an angle bisector)
(3) `/_A` is supplementary to `/_ADE` (Same Side Interior Angles)
So `m/_A+15x-11=180` or `m/_A=-15x+191`
(4) `/_ACF` is an exterior angle of `Delta ABC` , so its measure is equal to the sum of the remote interior angles. Thus:
(6) `m/_DBE=m/_BDE=4x-7=21` .
So `Delta ABD` is an acute, scalene triangle. (All angles less than 90, and no sides the same length.