Find area of triangle given one angle and the lengths its altitude divides the opposite side into

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In triangle $\triangle ADC$, $DB$ is perpendicular to $AC$ at $B$ so that $AB=2$ and $BC=3$ as shown in the figure. Furthermore, $\angle ADC=45^\circ$ . Find the area of $\Delta ADC$.

Tried the following things:

  1. Drawing medians and altitudes of the right triangles formed.
  2. Apollonius's theorem.
  3. Stewart's theorem.
  4. Sine rule.
  5. Cosine rule.
  6. Noticing that $3+2$ equals to $5$, I extended segment $BD$ to form a 3-4-5 triangle and try to proceed.

Spent around 1.5 hours on it. In each of the cases, I wasn't able to proceed after a certain point.

Best Answer

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$1=\tan(\alpha + \beta) = {\tan(\alpha)+\tan(\beta)\over 1-\tan(\alpha)\tan(\beta)}$

${\tan(\alpha) \over \tan(\beta)} = {2\over 3}$

Solve them you get $\tan(\beta)={1\over 2}$ so the height is $6$

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