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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