Answer:
Solution given;
<ABD=<BAC+<ACB
Since exterior angle of a triangle is equal to the sum of two opposite interior angle
26x+20=19x-15+9x+25
solve like terms
26x+20=28x+10
subtracting both by 10
26x+20-10=28x+10-10
Subtracting both side by 26x
10=28x-26x
2x=10
dividing both side by 2
2x/2=10/2
x=5
Now
<ABD=26*5+20=l50°
The value of <ABD is 150°