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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.

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  • $\begingroup$ I know one way tedious but effective is to try the problem in coordinate geometry label the vertices with coordinates and find coordinates of F, G and H(assume E is origin) and then prove that they lie on a circle. Though this is very inelegant and really a brute force way. This site mentions the precise coordinates for incentre- "mathopenref.com/coordincenter.html", I tried but it seemed very tedious this way. $\endgroup$ – Mustang Dec 22 '18 at 20:27
  • $\begingroup$ are you looking for pure euclidean approach, or is trigonometry allowed ? $\endgroup$ – G Cab Dec 23 '18 at 13:46
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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.

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  • $\begingroup$ that's beautiful! $\endgroup$ – Vendetta Dec 23 '18 at 14:39
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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.

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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.

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  • $\begingroup$ Thanks a lot! they are all great, though I gotta say I love first proof the best, particularly that little lemma. :) $\endgroup$ – Vendetta Dec 24 '18 at 20:07

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