*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)

`m/_ADE=11x-4+4x-7=15x-11`

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:

`17x+9=8x-14+m/_A`

So `m/_A=9x+23`

(5) `-15x+191=9x+23`

`-24x=-168`

`x=7`

(6) `m/_DBE=m/_BDE=4x-7=21` .

Then `m/_BED=180-42=138`

(7) `m/_A=9x+23=86,m/_ABD=4x-7=21,m/_ADB=11x-4=73`

So `Delta ABD` is an acute, scalene triangle. (All angles less than 90, and no sides the same length.

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