Triangle $ABC$ is inscribed in circle $(O)$, another circle $(O')$ touches $AB,AC$ at $P, Q$ respectively and touches circle $(O)$ internally at $S$. The lines $SP, SQ$ intersect $(O)$ at orther points $M, N$. Points $E, D, F$ are the perpendicular feet of point $S$ on $AM, MN, NA$ respectively. Prove that $DE=DF$

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Would you by any chance have a diagram? – user60887 Jul 9 '13 at 3:50

This problem is a combination of the two well-known facts:

Lemma 1. $M$ is the midpoint of the arc $AB.$

Proof. Homothety, with the center $S$ maps the small circle into a big one. So the tangent $AB$ to the small circle must be parallel to the tangent at $M$ to the big one. This implies the result.

Therefore, both $M$ and $N$ are the midpoints of the corresponding arcs.

Lemma 2. The center $I$ of the incircle of $\triangle ABC$ coincides with the midpoint of $PQ$.

Proof. Note, that the same homothety maps $\triangle SQP$ into $\triangle SNM$ so $MN\ ||\ QP$. Let $CM$ meet $PQ$ at point $I'.$ Then, $\angle PQS=\angle MNS=\angle MCS=\angle I'CS,$ so $I',Q,C,S$ lie on the same circle and $\angle QSI'=\angle QCI'=\frac{C}{2}.$ Thus, $\angle I'SP=\angle QSP-\angle I'SQ=\angle AQP-\frac{C}{2}=\frac{B}{2}$ and $\angle PI'S=\angle QCS=180^\circ-\angle PBS,$ so $I',P,B,S$ lie on the same circle. Thus, $\angle I'BP=\angle I'SP=\frac{B}{2}$ and $BI'$ is a bisector of $B.$ Therefore, $I'=I$ and $AI$ is a bisector of $\angle BAC.$ This implies that $I$ is a midpoint of $QP.$

Observe that the areas of $QIS$ and $PIS$ are the same so $\frac{1}{2}QS\cdot SI'\sin\angle QSI=\frac{1}{2}PS\cdot SI'\sin\angle PSI$ or $$\frac{SQ}{SP}=\frac{\sin{\frac{\angle B}{2}}}{\sin{\frac{\angle C}{2}}}.$$

Now, $F,D,N,S$ belong to the circle with diameter $SN,$ so $FD=SN\sin\angle{FND}=SN\sin\angle{ANM}=SN\sin{\frac{\angle C}{2}}.$ By analogy, $DE=SM\sin{\frac{\angle B}{2}}.$ So we are left to show that $$\frac{SN}{SM}=\frac{\sin{\frac{\angle B}{2}}}{\sin{\frac{\angle C}{2}}}.$$

It is left to note that $\triangle SQP$ is similar to $\triangle SNM$ and
$$\frac{SN}{SM}=\frac{SQ}{SP}.$$

Remark: Pure synthetic approach is also possible. All "trigonometric" conclusions could be deduced from the similarities of the corresponding triangles.

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