Two circumferences and the incircle

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Two circumferences and the incircle

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#1 h Dec 22, 2017, 1:05 AM • 2 Y

Y by Adventure10, Mango247

The incircle of a triangle scalene $ABC$ with center $I$ , touch the sides $BC$ and $CA$ in $D$ and $E$ respectively. Denote with $H$ the orthocenter of $ABI$ . $K=AI\cap BH$ and $L=BI\cap AH$ . Prove that the circumcircles of the triangles $DKH$ and $ELH$ they are cut at a point of the incircle of $ABC$ .

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#2 h Dec 22, 2017, 2:26 AM • 1 Y

Y by Adventure10

Any ideas?

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#3 h Dec 22, 2017, 3:00 AM • 3 Y

Y by macsnegro18, Adventure10, Mango247

Let $F$ be the point where the incircle touches $AB$ . Let $G$ be the intersection of $FI$ and the incircle. I claim that quadrilaterals $ELGH$ and $DKGH$ are cyclic.

Note that $\angle LHG=90-\angle LIH=90-\angle FIB=\frac{B}{2}$

Now, we want to find $\angle LEG$ . Note that $\angle LEG=90-\angle IEL-\angle CEG$

Since $CE$ is tangent to the incircle, $\angle CEG=\angle GFE$ , and since $EILFA$ is cyclic, $\angle EFI=\angle EAI=\frac{A}{2}$

Also because $EILFA$ is cyclic, $\angle LEI=\angle LAI=\angle KBI$
Consider the reflection of $B$ across $K$ . This lies on AC, so $\angle KBC+\angle C=\angle AB'K=\angle ABK=\angle B-\angle KBC$ . This implies that $\angle KBC=\frac{B-C}{2}$ , so $\angle IEL=\frac{B}{2}-\angle KBC=\frac{C}{2}$

Now, we have $\angle LEG=90-\frac{C}{2}-\frac{A}{2}=\frac{B}{2}$

Thus, $\angle LEG=\angle LHG$ , so $L, E, H, G$ are concyclic. Similarly, we can find that $D, K, H, G$ are concyclic, which solves the problem.

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2016 posts

PROF65

#5 h Dec 22, 2017, 6:30 PM • 2 Y

Y by Adventure10, Mango247

cosider the circles $C_1=(AEF),C_2=(BFD), C_3=(DHK),C_4=(EHL)$ we have $C_1\cap C_2=\{ F,I\},C_3\cap C_2=\{ D,K\},C_3\cap C_4=\{ H,P\},C_1\cap C_4=\{ E,L\}$ ;since the first bi-ntersections I,K,L and H are concyclic then using the theorem of six circles the second bi-intersections are also cyclic so $P$ is on $(I) \ \ \square .$
RH HAS

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91 posts

#6 h Dec 22, 2017, 9:19 PM • 1 Y

Y by Adventure10

PROF65 wrote:

cosider the circles $C_1=(AEF),C_2=(BFD), C_3=(DHK),C_4=(EHL)$ we have $C_1\cap C_2=\{ F,I\},C_3\cap C_2=\{ D,K\},C_3\cap C_4=\{ H,P\},C_1\cap C_4=\{ E,L\}$ ;since the first bi-ntersections I,K,L and H are concyclic then using the theorem of six circles the second bi-intersections are also cyclic so $P$ is on $(I) \ \ \square .$
RH HAS

Could you please explain the theorem you used ?

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jlammy

#7 h Dec 22, 2017, 9:56 PM • 1 Y

Y by Adventure10

Let the incircle touch $AB$ at $F$ , and let $G$ be diametrically opposite $F$ ; we claim that $G$ lies on $(DKH)$ and $(ELH)$ .

It's well-known that $K$ , $L$ lie on $DE$ (see Yufei Zhao Lemma 8). Then $\angle GDK=90^{\circ}-\angle EDF=\tfrac{1}{2}\angle A=\angle GHK$ , so $G$ lies on $(DKH)$ , and analogously $G$ lies on $(ELH)$ .

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2016 posts

PROF65

#8 h Dec 22, 2017, 10:21 PM • 2 Y

Y by Adventure10, Mango247

omarius wrote:

PROF65 wrote:

cosider the circles $C_1=(AEF),C_2=(BFD), C_3=(DHK),C_4=(EHL)$ we have $C_1\cap C_2=\{ F,I\},C_3\cap C_2=\{ D,K\},C_3\cap C_4=\{ H,P\},C_1\cap C_4=\{ E,L\}$ ;since the first bi-ntersections I,K,L and H are concyclic then using the theorem of six circles the second bi-intersections are also cyclic so $P$ is on $(I) \ \ \square .$
RH HAS

Could you please explain the theorem you used ?

just a diagram

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