+130071 This might not be the easiest approach....but.....
By Leg - Leg , the two right triangles are congruent.....so DC = DB =11
Since angle ABC is bisected, then AB =(25/11)BC
And, since angles ADB and CDB are supplementary the cosine of ADC = - cosCDB
By the Law of Cosines
( BC )^2 = 11^2 + 11^2 - 2 (11)(11) (cos CDB)
(BC)^2 = 242 - 242 ( cos CBD)
[(BC)^2 - 242 ] / [ -242] = ( cos CBD ) (1)
And
[ (25/11)BC]^2 = 11^2 + 25^2 - 2 (11)(25) (-cos CBD)
[ (625 /121)BC^2 ] = 746 + 550 (cos CBD)
[ (625/121)BC^2 - 746 ] / [550] = (cos CBD) (2)
Equate ( 1) and (2)
[ (BC)^2 - 242 ] / [ -242] = [ (625 /121) BC^2 - 746 ] / 550 cross-multiply
550 [ (BC)^2 - 242 ] = (-242)[ (625/ 121) (BC^2) - 746] simplify
550BC^2 - 133100 = -1250BC^2 + 180532
1800BC^2 = 313632
BC^2 = 313632 / 1800
BC = sqrt [ 313632 / 1800 ] = 13.2
So AB = (25/11) (13.2) = 30
The semi-perimeter of ABD = (25 + 11 + 30) / 2 = 33
So ....using Heron's Formula, [ ABD ] is
sqrt [ 33 * ( 33-25) * (33 - 11) * ( 33-30) ] =
sqrt [ 33 * 8 * 22 * 3] =
sqrt [ 17424 ] =
132 units ^2
![$[ABD].$](/api/ssl-img-proxy?src=https%3A%2F%2Flatex.artofproblemsolving.com%2F0%2F6%2F9%2F06983453b462c22335a6d642fc655fac8968d328.png)
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