<s>[INST] Right angled triangle and its incircle $ABC$ is a right-angled triangle ($\measuredangle ACB=90^\circ$) and $O$ is its incenter. If $BO=m$ and $\measuredangle BAC=\alpha$, find the area of the triangle $ABC$.


 the given answer is $$\dfrac{m^2}{2}\cos\alpha\cot\dfrac{\alpha}{2}$$

Let $OP\perp AB, OP=r$. As $\measuredangle ABC=90^\circ-\alpha$, then $\measuredangle OBP=45^\circ-\dfrac{\alpha}{2}$ and $$BP=m\cos(45^\circ-\alpha)\\AP=r\cot\dfrac{\alpha}{2}\\r=m\sin\left(45^\circ-\dfrac{\alpha}{2}\right)$$ (I have used $r$ for simpler calculations.) Because of the equal tangent segments, the semiperimeter is $$p=r\cot\dfrac{\alpha}{2}+m\cos\left(45^\circ-\dfrac{\alpha}{2}\right)+r$$ Now putting $r=m\sin\left(45^\circ-\dfrac{\alpha}{2}\right)$ gives $$p=m\sin\left(45^\circ-\dfrac{\alpha}{2}\right)\cot\dfrac{\alpha}{2}+m\cos\left(45^\circ-\dfrac{\alpha}{2}\right)+m\sin\left(45^\circ-\dfrac{\alpha}{2}\right)\\=m\left(\sqrt2\cos\dfrac{\alpha}{2}+\sin\left(45^\circ-\dfrac{\alpha}{2}\right)\cot\dfrac{\alpha}{2}\right)$$ using $$\sin\left(45^\circ-\dfrac{\alpha}{2}\right)+\cos\left(45^\circ-\dfrac{\alpha}{2}\right)\\=\sin\left(45^\circ-\dfrac{\alpha}{2}\right)+\sin\left(45^\circ+\dfrac{\alpha}{2}\right)=2\sin45^\circ\cos\dfrac{\alpha}{2}$$
I don't really see how to reach the given answer. Any help would be appreciated.
 [/INST] The area of $\Delta ABC$ is $r(PB+AP)+r^2$ (adding area of the square to the sum of areas of the two quadrilaterals). Pythagorus theorem on $\Delta ABC$ gives that this is just $PB.AP$. Now $PB=r \cot(\pi/4 - \alpha/2)$ and $AP=r\cot(\alpha/2)$. Again, from $r^2+PB^2=m^2$, get $r^2=m^2\sin^2(\pi/4-\alpha/2)$. Now put everything together and expand it out a bit, things cancel and you get what you want.
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