幾何問題求解 ?

<DAB = <ACB (given)
<DBA = <ABC (given)
<ADB = <CAB (< sum of △)

so, △DAB ~ △ACB (equiangular)

AB/DB = CB/AB (corr. sides,~△)
36/DB = CB/36
36^2 = DB(2DB) (AD為中線)
1296 = 2DB^2
DB = 18√2

tan<DAB = DB/AB = 18√2 / 36
<DAB=35.3°
<CAB=180°-35.3°-90°(< sum of △)
=54.7°

cos<CAB =AE/AB
AE=36(cos54.7°)
=20.8

<EAF=180°-35.3°-90°-35.3°(< sum of △)
=19.5°

tan<EAF= EF/AE
tan19.5°= EF/20.8
EF= 7.35

EF^2=7.35^2
=54

finished