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### incircles and excircles

Posted: Fri Jul 10, 2015 12:52 am
Given a point $M$ on the side $AB$ of the triangle $\Delta ABC$ , let ${r}_{1},{r}_{2}$ are the radii of the inscribed circles of the triangles $\Delta ACM,\Delta BCM$ respectively and let ${p}_{1},{p}_{2}$ are the radii of the excircles of the triangles $\Delta ACM,\Delta BCM$ at the sides $AM$ and $BM$ respectively . let $r,p$ denote the radii of incircle and excircle respectively of $\Delta ABC$. prove that , $\frac{{r}_{1}}{{p}_{1}}\frac{{r}_{2}}{{p}_{2}}=\frac{r}{p}$

### Re: incircles and excircles

Posted: Mon Apr 02, 2018 12:16 pm
Let $R_1$ and $P_1$ be the incenter and the C-excenter of triangle $AMC$.
$R_2$ and $P_2$ be the incenter and the C-excenter of triangle $BMC$.
So, we get $AP_1MR_1$ and $BP_2MR_2$ cyclic.
$$\angle{AMR_1}=\angle{R_1MC}=90^o-\angle{R_2MC} \angle{R_2MC}=\angle{R_2MB} \angle{AMP_1}=90^o-\angle{R_1MC}$$
We can easily show that, $$\frac{\sin\angle{AMR_1}}{\sin\angle{AMP_1}}=\tan\angle{R_1MC}$$ and $$\frac{\sin\angle{BMR_2}}{\sin\angle{BMP_2}}=\tan\angle{R_2MC}$$
$$\frac{r_1}{p_1}\frac{r_2}{p_2}=\frac{r_1}{MR_1}\frac{MP_1}{p_1}\frac{MR_1}{MP_1}\frac{r_2}{MR_2}\frac{MP_2}{p_2}\frac{MR_2}{MP_2}=\frac{\sin\angle{AMR_1}}{\sin\angle{AMP_1}}\tan\frac{1}{2}\angle{A}.\frac{\sin\angle{BMR_2}}{\sin\angle{BMP_2}}\tan\frac{1}{2}\angle{B}=\tan\angle{R_1MC}.\tan\angle{R_2MC}.\tan\frac{1}{2}\angle{A}.\tan\frac{1}{2}\angle{B}=\tan\frac{1}{2}\angle{A}.\tan\frac{1}{2}\angle{B}$$
Now, $$\frac{r}{p}=\frac{r}{AI}.\frac{AI_C}{p}.\frac{AI}{AI_C}=tan\frac{1}{2}\angle{A}.tan\frac{1}{2}\angle{B}$$ can easily be shown......