What do you mean by R? Surely not the circumradius? darij

Possibly he means the inradii $r_{b},\ r_{c}$ of $\triangle PBD,\ \triangle PCD$ are equal. Let $s_{b}, s_{c}$ be the semiperimeters and $\triangle_{b},\ \triangle_{c}$ areas of these 2 triangles. Their D-altitude is common, hence their areas are in the ratio of their bases. $\frac{r_{b}}{r_{c}}= \frac{\triangle_{b}}{\triangle_{c}}\cdot \frac{s_{c}}{s_{b}}= \frac{BP}{CP}\cdot \frac{s_{c}}{s_{b}}$ ABDC is cyclic, $BP \cdot CP = AP \cdot DP$ and triangles $\triangle ABP \sim \triangle CDP$ are similar, $\frac{AB}{AP}= \frac{CD}{CP},$ and $AB \cdot CP = AP \cdot CD.$ Adding the 2 products, $(AB+BP) \cdot CP = (CD+DP) \cdot AP$ AP is the perimeter splitter of $\triangle ABC$ from A, $AB+BP = AC+CP.$ Substituting this, $(AC+CP) \cdot CP = (CD+DP) \cdot AP$ Adding $AP \cdot CP$ to both sides, $(AC+AP+CP) \cdot CP = (CP+CD+DP) \cdot AP$ $\frac{AC+AP+CP}{AP}= \frac{CP+CD+DP}{CP}$ Triangles $\triangle PBD \sim \triangle PAC$ are similar with similarity coefficient $\frac{BP}{AP}$ and so are their perimeters, $\frac{BD+BP+DP}{AC+AP+CP}= \frac{BP}{AP}$ Substituting this to the previous equation, $\frac{BD+BP+DP}{BP}= \frac{CP+CD+DP}{CP}$ $\frac{s_{b}}{BP}= \frac{s_{c}}{CP},\ \ \frac{BP}{CP}\cdot \frac{s_{c}}{s_{b}}= 1$ and it follows that the inradii $r_{b}= r_{c}$ are equal.

Attachments:

Omid Hatami wrote: The $A$- excircle of the triangle $ABC$ is tangent to the side $[BC]$ at the point $P$. The line $AP$ intersects again the circumcircle of the triangle $ABC$ at the point $D$. Prove that the inradius of the triangles $BDP$ and $CDP$ are equally. Proof (in the my opinion, although alike with Yetti's, it is a bit more clearly). For the triangle $XYZ$ denote the semiperimeter $s(XYZ)$ and the area $[XYZ]\ .$ $\{\begin{array}{ccc}AB+BP=AC+CP & \Longrightarrow & s(PAB)=s(PAC)\\\ PCD\sim PAB & \Longrightarrow & \frac{s(PCD)}{s(PAB)}=\frac{PC}{PA}\\\ PAC\sim PBD & \Longrightarrow & \frac{s(PAC)}{s(PBD)}=\frac{PA}{PB}\end{array}$ $\Longrightarrow$ $\frac{[PDB]}{[PDC]}=\frac{PB}{PC}=\frac{r_{1}}{r_{2}}\cdot\frac{s(PBD)}{s(PCD)}$ $\Longrightarrow$ $\frac{r_{1}}{r_{2}}=\frac{PB}{PC}\cdot\frac{s(PCD)}{s(PBD)}=\frac{PB}{PC}\cdot\frac{(PCD)}{(PAB)}\cdot\frac{(PAB)}{(PAC)}\cdot\frac{(PAC)}{(PBD)}=\frac{PB}{PC}\cdot\frac{PC}{PA}\cdot 1\cdot\frac{PA}{PB}=1$ $\Longrightarrow$ $\boxed{\ r_{1}=r_{2}\ }\ .$ Remark. See http://www.mathlinks.ro/Forum/viewtopic.php?t=50559 . Generally, for any point $P\in (BC)$ there is the relation $\boxed{\ \frac{r_{1}}{r_{2}}=\frac{s(PAB)}{(PAC)}\ }\ !$ Can somebody ascertain $r_{1}$ only in accordance with the elements of the triangle $ABC\ ?$ Yetti, now in the your proof you didn't use the remarkable relation $\frac{1}{r_{1}}+\frac{1}{r_{2}}=\frac{r}{r_{1}r_{2}}+\frac{2}{h}\ .$

With the yetti's notations, besides denote $r_{1},r_{2}$ inradii of $ABP,APC$. Now by areas: $\frac{[ABP]}{[APC]}=\frac{r_{1}}{r_{2}}=\frac{BP}{PC}=\frac{[PBD]}{[PDC]}$ ..(1) $[ABP]=\frac{r_{1}^{2}}{r_{c}^{2}}[PDC]$ ..(2) $[APC]=\frac{r_{2}^{2}}{r_{b}^{2}}[PBD]$ ..(3) Combinig (2) and (3) with (1): $\frac{[ABP]}{[APC]}=\frac{r_{1}^{2}r_{b}^{2}[PDC]}{r_{2}^{2}r_{c}^{2}[PBD]}$ $\frac{r_{1}}{r_{2}}=\frac{r_{1}^{2}r_{b}^{2}r_{2}}{r_{2}^{2}r_{c}^{2}r_{1}}$ $r_{b}=r_{c}$.