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Find $[ABD].$

[asy] unitsize(1.2 cm);pair A, B, C, D, M;D = (0,0);C = (2,0);B = 2*dir(80);A = extension(C, D, B, reflect(B,D)*(C));M = (B + C)/2;draw(A--B--C--cycle);draw(B--D--M);draw(rightanglemark(B,M,D,5));add(pathticks(B--M, 1, .5, 6, 8));add(pathticks(C--M, 1, .5, 6, 8));draw(anglemark(A,B,C,15));add(pathticks(anglemark(A,B,D,15),1,0.15));add(pathticks(anglemark(D,B,C,15),1,0.15));label(

 Apr 13, 2022
 #1
avatar+124592 
+1

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

 

 

cool cool cool

 Apr 13, 2022

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