In the right triangle $ \triangle AEI,$ $ \overline{AM}$ is the orthogonal projection of the leg $ \overline{AE}$ onto its hypotenuse $ \overline{AI}$ $ \Longrightarrow$ $ AE^2 = AM \cdot AI,$ but from the power of $ A$ with respect to the incircle $ (I),$ we have $ AE^2 = AP \cdot AD$ $ \Longrightarrow$ $ AM \cdot AI = AP \cdot AD$ $ \Longrightarrow$ Four points $ P,M,I,D$ are concyclic.

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Martin N. wrote: Let $ \triangle{ABC}$ be a triangle with $ AB\not = AC$. The incircle with centre $ I$ touches $ BC$, $ CA$, $ AB$ at $ D$, $ E$, $ F$, respectively. Furthermore let $ M$ the midpoint of $ EF$ and $ AD$ intersect the incircle at $ P\not = D$. Show that $ PMID$ ist cyclic. An inversion in the incircle will transform $ M$ into $ A$ and fix both $ P$ and $ D$. Since $ A,P,D$ are colinear the conclusion follows.

We have: $AE^2$ = $AF^2$ = $\overline{AM}$ . $\overline{AI}$ = $\overline{AP}$ . $\overline{AD}$ So: $P$, $M$, $I$, $D$ lie on a circle

Dear Mathlinkers, Iberoamerican Olympiad (1990) Problem 4 Sincerely Jean-Louis