simple looking but hard to prove geometrical problem: prove that 4 points on the same circle. Pardon my bad drawing. ABCD is a square. E is any point on CD. F,G,H are the incenters of triangles BCE, ABE and ADE. Prove that EFGH are on the same circle. 

 A: First, let's prove an intermediate conclusion, or a lemma, which can be stated as follows.

Lemma Let $l$ be another exterior common tangent (namely, not $CD$) of the
  circles $(ADE)$ and $(BCE)$. Then $l$ is tangent to the circle
  $(ABE)$.

Proof   All points are labeled as the figure shows. Notice that
\begin{align*}
AK&=AO-KO=\frac{1}{2}(AD+AE-DE)-KJ,\\
BL&=BN-LN=\frac{1}{2}(BC+BE-CE)-LM.\\
\end{align*}
Hence
\begin{align*}
AK+BL&=AB+\frac{1}{2}(AE+BE-DE-CE)-(KJ+LM)\\
&=AB+\frac{1}{2}(AE+BE-CD)-(JM-KL)\\
&=AB+\frac{1}{2}(AE+BE-CD)-(PQ-KL)\\
&=AB+KL+\frac{1}{2}(AE+BE-CD)-(EP+EQ)\\
&=AB+KL,\\
\end{align*}
which shows that the quadrilateral $ABLK$ has an inscribed circle. Apparently, it must be the one of triangle $AEB$, namely, $ABLK$ and $ABE$ have the identical inscribed circle. The conclusion we want to prove is followed. Moreover, we may see that, $AE,GH,l$ and $BE,GF,l$ are respectively concurrent.
Now, come back to deal with the present problem. Notice that $GA||EH$ and $GB||EF$. hence $\angle HEF=\angle AGB$. But $ABLK$ is a circumscribed quadrilateral, then it's obvious that $\angle AGB+\angle KGL=180^o$. As a result, $\angle HEF+HGF=180^o$, which implies that $H,E,F,G$ are concyclic. We are done.

A: Another Proof
Let $O$ be the intersection point of $AC$ and $BD$, and $X$ be the orthogonal projection of $G$ on $AB.$ It's obvious that $F,H$ lie on $AC,BD$ respectively, and $\angle AHE=\angle BFE=135^o.$
Observe the two triangles $BXG$ and $BOF$. We may see $$\angle BXG=\angle BOF=90^o,$$ and $$\angle XBG=\dfrac{1}{2}\angle ABE=45^o-\frac{1}{2}\angle CBE=\angle OBC-\angle FBC=\angle OBF.$$ Therefore $$\triangle BXG \sim \triangle BOF.$$Thus,$$\frac{BX}{BO}=\frac{BG}{BF}.$$ Futher $$\triangle BXO \sim \triangle BGF.$$It follows that $$\angle BOX=\angle BFG.$$
Likewise, $$\angle AOX=\angle AHG.$$
Hence
\begin{align*}
\angle EFG+\angle EHG&=(\angle EFB-\angle GFB)+(\angle EHA-\angle AHG)\\
&=(\angle EFB+\angle EHA)-(\angle BOX+\angle AOX)\\
&=(\angle EFB+\angle EHA)-\angle AOB\\
&=2\cdot 135^o-90^o\\
&=180^o,
\end{align*}
which implies $E,F,G,H$ are concyclic.

A: Three-Chord Theorem
Let $PABC$ be a quadrilateral. $P,A,B,C$ are concyclic if and only if
$$PB\cdot\sin \angle APC=PA\cdot\sin \angle BPC+PC \cdot\sin \angle APB.$$
This is nothing but a kind of transformation of well-known Ptolemy's Theorem, and hence we do not intend to give its proof.
Let's take up the present problem, applying the theorem above. It's easy to obtain that
$$EF\cdot \sin \angle GEH=EF \cdot \cos \angle BEF=\frac{EB+EC-BC}{2},$$
$$EH\cdot \sin \angle FEG=EH \cdot \cos \angle AEH=\frac{EA+ED-AD}{2},$$
$$EG\cdot \sin \angle FEH=EH \cdot \cos \angle AEG=\frac{EA+EB-AB}{2}.$$
Since
$$\frac{EA+EB-AB}{2}=\frac{EB+EC-BC}{2}+\frac{EA+ED-AD}{2},$$
then
$$EG\cdot \sin \angle FEH=EF\cdot \sin \angle GEH+EH\cdot \sin \angle FEG.$$
Accoring to Three-Chord Theorem, $E,F,G,H$ are concyclic. The proof is completed.
