Let ∠BAD=a, then ∠ DAC = ∠ a-a = 45-a.
Known: ∠ BDA = ∠ CDA = 90, BD=2 and CD=3.
△ in△ BDA: BD/AD = TANA
Namely: ad = BD/tana = 2/tana .......................... (1).
△In△CDA:CD/AD = Tan(45-a)=(Tan 45-Tana)/( 1+Tan 45 Tana)=( 1-Tana)/( 1+Tana)
That is: AD = CD( 1+ tana)/(kloc-0/-tana) = 3( 1+ tana) /( 1- tana) ......................................................... (2).
Yes: 2/ Tanner = 3( 1+ Tanner) /( 1- Tanner)
2( 1- tana) = 3 tana (1+ tana)
3(tana)^2+5tana-2=0
(3tana- 1)(tana+2)=0
Yes: 3tana- 1=0, tana+2=0.
Solution: tana= 1/3, tana=-2, negative values are discarded.
Substituting tana= 1/3 into (1) gives: AD=2/( 1/3)=6.
AD is the height of △ABC, BC=BD+CD=2+3=5.
The area of △ABC is: (1/2 )× BC× ad = (1/2 )× 5× 6 =15.
To sum up, there are:
AD=6, the area of △ABC is 15.
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