Online Course Discussion Forum
Centers of a Triangle
Re: Centers of a Triangle
Your question was messed up and I'll guess what you were asking: How did we get
$$ \frac{DM\cdot\sin\angle PDM}{DI\cdot\sin\angle PDI} = \frac{2R\sin^2\alpha}{2R\sin\alpha}\cdot\frac{\sin(90^\circ - \alpha +\alpha - \beta)}{\sin(C+\alpha-\beta)}? $$ To get this, we connect $DC$ and $DB$. In right $\triangle DCM$, $\angle MCD=\alpha$, and $DC=2R\sin\alpha$, thus $DM=2R\sin^2\alpha$. Also $DI=DC=2R\sin\alpha$. In right $\triangle BMD$, $\angle BDM=90^\circ - \alpha$, and $\angle BDN=\angle BAN=\alpha-\beta$, so $\angle PDM=90^\circ - \alpha + \alpha - \beta = 90^\circ - \beta$, and $\angle PDI =\angle NCA=C+\angle BCN = C+\alpha-\beta$. With these results, we get the equality above.
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