My solution. Let $CH$ cut $(O)$ at $R$, $AI$ cut $(O)$ at $T$. Applying Pascal's theorem for $(AARBTC)$, we gets that $R,K,T$ are collinear. Note that $KR=KH$, so we have $\angle AKH=\angle RKA$, but since $\angle RKA+\angle TRC=\angle RKA+\angle BAC.\frac {1}{2}$, so we have $KH\perp AD.$

Let $CH\cap AB=S$. From menalaus theorem for $P,K,D$ in $\triangle BSC$ we get: $\frac{KS}{KB}=\frac{DC}{BD}\times \frac{PS}{PC}=\frac{b}{c}\times \frac{AS}{AC}\times \frac{\sin C}{\sin B}=\frac{AS}{AC}=\frac{HS}{HK}\Longrightarrow HK$ is bisector of $\angle SHB\Longrightarrow \angle HKA=90-\frac{A}{2}$. Q.E.D

Overly long purely synthetic solution. For the sake of brevity, denote by $a$ the line tangent to the circumcircle of triangle $ABC$ at the vertex $A$. Let the $B$ altitude meet $a$ again at $Q$. Let $DP \cap AB=K'$ and $DQ\cap AC=L'$. By Pappus' theorem on $(CDB,QAP)$ we conclude that the points $L',H,K'$ are collinear. Let the point $T$ be the harmonic conjugate of the feet $F$ of the $A$ altitude. Note that $D(L',K';H,\infty)=(Q,P;A,X)$ for some point $X$ on $a$ such that $\angle ADX=90^{\circ}$. Moreover, $-1=H(B,C;F,T)=(Q,P;A,X')$ for $X'$ as the intersection of $a$ with $TH$. We need to show that $X=X'$. Let $M$ be the midpoint of $BC$ and $N$ be the midpoint of arc $BC$ of the circumcircle of triangle $ABC$. Let $O$ be the centre of the circumcircle. It is well-known that $TH \perp AM$, $O,M,N$ are collinear and $A,D,N$ are collinear. Therefore, we have the newly formulated problem: Points $H$, $D$ and $M$ are the orthocenter, the feet if $A$ angle bisector and the midpoint of side $BC$, respectively in triangle $ABC$. Prove that the line $a$, the perpendicular from $D$ to $AD$ and the perpendicular from $H$ to $AM$ are concurrent. Proof: Consider an inversion about $A$ of radius $\sqrt{AH\cdot AF}$. This sends the line from $H$ perpendicular to $AM$ to the circle with diameter $AM$. Therefore, $X$ is mapped to $Y \in a$ such that $\angle AYM=90^{\circ}$. Consider an inversion about $A$ of radius $\sqrt{AB\cdot AC}$. This sends the line through $D$ perpendicular to $AD$ to the circle with diameter $AN$. Therefore, $X'$ is mapped to $Y' \in a$ such that $\angle AY'N=90^{\circ}$. Notice that the former inversion is simply a scaling of the latter by a factor of $\cos A$. Therefore, proving that $AY'=AY\cos A$ is sufficient. However, notice that $AY$ is the projection of $OM$ onto line $a$ and $AY'$ is the projection of $ON$ on line $a$. Since $OM=ON\cos A$, our conclusion holds.

Define $\{A, L\} \equiv AD \cap \odot(ABC)$, $\{C, F\}\equiv CH\cap \odot(ABC)$. By Pascal on $AABCFL$, we get that $K \equiv AB\cap FL$. Note that since $AB$ is the perpendicular bisector of $\overline{HF}$, $KH=KF$ and so $$\angle AKH=\frac{\angle FKH}{2}=\frac{\pi-2\angle HFK}{2}=\frac{\pi-2\angle CAL}{2}=\frac{\pi-\angle A}{2}$$so with $E\equiv KH\cap AC$, we get that $\triangle AKE$ is isosceles, as desired.

doxuanlong15052000 wrote: My solution. Let $CH$ cut $(O)$ at $R$, $AI$ cut $(O)$ at $T$. Applying Pascal's theorem for $(AARBTC)$, we gets that $R,K,T$ are collinear. Note that $KR=KH$, so we have $\angle AKH=\angle RKA$, but since $\angle RKA+\angle TRC=\angle RKA+\angle BAC.\frac {1}{2}$, so we have $KH\perp AD.$ Maybe some mistake in express

Dear Mathlinkers, by applying a second time Pascal's theorem, H appears to be the orthocenter of another triangle and we are done... Sincerely Jean-Louis

We will barycentric. $H=(S_{BC}:S_{CA}:S_{AB})$ $D=(0:1:1)$ From The line tangents the circumcircle at A.We get equation $c^2y+b^2z=0$ So $P=(b^2S_{B}:b^2S_{A}:-c^2S_{A}) \implies P=(\frac{b^2S_{B}}{b^2S_{B}+b^2S_{A}-c^2S_{A}}:\frac{b^2S_{A}}{b^2S_{B}+b^2S_{A}-c^2S_{A}}:\frac{-c^2S_{A}}{b^2S_{B}+b^2S_{A}-c^2S_{A}}) $ Let $K=(k:1-k:0)$ K line on PD. SO $ \begin{vmatrix} 0 & 1 &1 \\ b^2S_{B}& b^2S_{A} & -c^2S_{A} \\ k& 1-k & 0 \end{vmatrix} =0 $ We get $K=(\frac{b^2S_{B}}{b^2S_{B}+b^2S_{A}+c^2S_{A}},\frac{b^2S_{A}+c^2S_{A}}{b^2S_{B}+b^2S_{A}+c^2S_{A}},0)=(\frac{b^2S_{B}}{b^2c^2+c^2S_{A}},\frac{b^2S_{A}+c^2S_{A}}{b^2c^2+c^2S_{A}},0)$ $\overrightarrow{KH}=(S_{A}(S_{C}(b^2c^2+c^2S_{A})-(b^2+c^2)S^2)),S_{A}(S_{C}(b^2c^2+c^2S_{A})-(b^2+c^2)S^2),S_{CA}S^2(b^2c^2+c^2S_A))$ $\overrightarrow{AD}=(2,-1,-1)$ $a^2(S_{CA}S^2(b^2c^2+c^2S_A)(-1)+S_{A}(S_{C}(b^2c^2+c^2S_{A})-(b^2+c^2)S^2)(-1))+b^2(S_{A}(S_{C}(b^2c^2+c^2S_{A})-(b^2+c^2)S^2))(-1)+S_{A}(S_{C}(b^2c^2+c^2S_{A})-(b^2+c^2)S^2))(2))+c^2(S_{A}(S_{C}(b^2c^2+c^2S_{A})-(b^2+c^2)S^2)(2)+S_{A}(S_{C}(b^2c^2+c^2S_{A})-(b^2+c^2)S^2))(-1))$ if It is not wrong the last equation should equal to zero.I waits someone to help me.

MonsterS wrote: We will barycentric. $H=(S_{BC}:S_{CA}:S_{AB})$ $D=(0:1:1)$ From The line tangents the circumcircle at A.We get equation $c^2y+b^2z=0$ So $P=(b^2S_{B}:b^2S_{A}:-c^2S_{A}) \implies P=(\frac{b^2S_{B}}{b^2S_{B}+b^2S_{A}-c^2S_{A}}:\frac{b^2S_{A}}{b^2S_{B}+b^2S_{A}-c^2S_{A}}:\frac{-c^2S_{A}}{b^2S_{B}+b^2S_{A}-c^2S_{A}}) $ Let $K=(k:1-k:0)$ K line on PD. SO $ \begin{vmatrix} 0 & 1 &1 \\ b^2S_{B}& b^2S_{A} & -c^2S_{A} \\ k& 1-k & 0 \end{vmatrix} =0 $ We get $K=(\frac{b^2S_{B}}{b^2S_{B}+b^2S_{A}+c^2S_{A}},\frac{b^2S_{A}+c^2S_{A}}{b^2S_{B}+b^2S_{A}+c^2S_{A}},0)=(\frac{b^2S_{B}}{b^2c^2+c^2S_{A}},\frac{b^2S_{A}+c^2S_{A}}{b^2c^2+c^2S_{A}},0)$ $\overrightarrow{KH}=(S_{A}(S_{C}(b^2c^2+c^2S_{A})-(b^2+c^2)S^2)),S_{A}(S_{C}(b^2c^2+c^2S_{A})-(b^2+c^2)S^2),S_{CA}S^2(b^2c^2+c^2S_A))$ $\overrightarrow{AD}=(2,-1,-1)$ $a^2(S_{CA}S^2(b^2c^2+c^2S_A)(-1)+S_{A}(S_{C}(b^2c^2+c^2S_{A})-(b^2+c^2)S^2)(-1))+b^2(S_{A}(S_{C}(b^2c^2+c^2S_{A})-(b^2+c^2)S^2))(-1)+S_{A}(S_{C}(b^2c^2+c^2S_{A})-(b^2+c^2)S^2))(2))+c^2(S_{A}(S_{C}(b^2c^2+c^2S_{A})-(b^2+c^2)S^2)(2)+S_{A}(S_{C}(b^2c^2+c^2S_{A})-(b^2+c^2)S^2))(-1))$ if It is not wrong the last equation should equal to zero.I waits someone to help me. Your solution is wrong.Because $D$ isn't midpoint of $ BC.$

Extend $BH$ cut $AP$ at $Q$ and let $L$ be the intersection point between $AC$ and $QD$. We know that $K,H,L$ are collinear points from Pappus' theorem. Outline of my remaining step 1. Show that $B,P,Q,C$ lie on the same circle. 2. Show that the circumcircle of $\triangle PQD$ touch $BC$ at $D$. 3. Comparing angles in the figure to show that $\angle KDA=\angle LDA$. Finally, we have $\triangle KDA \cong \triangle LDA$ and hence $KA=AL$. This implies that $KH \perp AD$.

Attachments:

doxuanlong15052000 wrote: My solution. Let $CH$ cut $(O)$ at $R$, $AI$ cut $(O)$ at $T$. Applying Pascal's theorem for $(AARBTC)$, we gets that $R,K,T$ are collinear. Note that $KR=KH$, so we have $\angle AKH=\angle RKA$, but since $\angle RKA+\angle TRC=\angle RKA+\angle BAC.\frac {1}{2}$, so we have $KH\perp AD.$ I don't get how to apply pascal's theorem

Skravin wrote: doxuanlong15052000 wrote: My solution. Let $CH$ cut $(O)$ at $R$, $AI$ cut $(O)$ at $T$. Applying Pascal's theorem for $(AARBTC)$, we gets that $R,K,T$ are collinear. Note that $KR=KH$, so we have $\angle AKH=\angle RKA$, but since $\angle RKA+\angle TRC=\angle RKA+\angle BAC.\frac {1}{2}$, so we have $KH\perp AD.$ I don't get how to apply pascal's theorem Pascal states that P,D, RT $\cap$ AB are collinear. But we know PD $\cap$ AB = K, so it follows that K = RT $\cap$ AB, and R, K, T are collinear.

vsathiam wrote: Skravin wrote: doxuanlong15052000 wrote: My solution. Let $CH$ cut $(O)$ at $R$, $AI$ cut $(O)$ at $T$. Applying Pascal's theorem for $(AARBTC)$, we gets that $R,K,T$ are collinear. Note that $KR=KH$, so we have $\angle AKH=\angle RKA$, but since $\angle RKA+\angle TRC=\angle RKA+\angle BAC.\frac {1}{2}$, so we have $KH\perp AD.$ I don't get how to apply pascal's theorem Pascal states that P,D, RT $\cap$ AB are collinear. But we know PD $\cap$ AB = K, so it follows that K = RT $\cap$ AB, and R, K, T are collinear. thanks

Many of solutions posted above undoubtedly overleap the case when $\triangle ABC$ is obtuse. Fortunately, jayme's brilliant idea would deal with all the cases.

Let $AD \cap PC=X$.By ratio Lemma on $\triangle APC$ with transversal $AX$ and $\triangle APD$ with transversal $AK$ we get that: $\frac{PK}{KD}=\frac{AP}{AD} . \frac{\sin C}{\sin \frac{A}{2}},\frac{PX}{XC}=\frac{AP}{AC} . \frac{\sin C+\frac{A}{2}}{\sin \frac{A}{2}} \Rightarrow \frac{PK}{KD}=\frac{PX}{XC}.\frac{AC}{AD} . \frac{\sin C}{\sin \frac{A}{2}+C} \Rightarrow \frac{PK}{KD}=\frac{PX}{XC}$ The last relation follows from sin law in $\triangle ADC$.So we have $BD||KX$ and so $AH \bot KX$ and $H$ will be the orthocenter of $\triangle AKX$ and we get $KH \bot AD$ so we are done.

Dear Mathlinkers, http://jl.ayme.pagesperso-orange.fr/Docs/Orthique%20encyclopedie%204.pdf p. 48-50. Sincerely Jean-Louis

See my solution to this problem on my Youtube channel here: https://www.youtube.com/watch?v=oXpKPj-PPo0

If we get $ K' $ on $ AB $ such that $ HK' \perp AD $ and $H'$ is the foot of $A$ on $BC$ , It's easy to check that $ A(PH,K'D) = H(PA,K'D) $ ( by using : $ A(PH,K'D) = \frac {\sin(C).\sin(\frac{B-C}{2})}{\cos(B).\sin(\frac{A}{2}+C)}$ and $ H(PA,K'D) = \frac {\sin(\frac{A}{2}).\sin(\angle{DHH'})}{\sin(\frac{A}{2}+C).\sin(\angle{DHC})}$ ) , So $ P , K' , D $ are collinear .

let $AD \cap (ABC)=M$ $CH \cap AD= S$ $CH \cap (ABC)=H'$ claim(1): $H',K,M$ are collinear proof: just pascal on $AAMH'CB$ $\blacksquare$ so $\angle SH'K=SH'M=CAM=BAM \implies ASKH'$ is cyclic so $SK \parallel BC$ so $AH \perp SK$ combined with $SH \perp Ak \implies KS \perp AD$

jrpartty wrote: Extend $BH$ cut $AP$ at $Q$ and let $L$ be the intersection point between $AC$ and $QD$. We know that $K,H,L$ are collinear points from Pappus' theorem. Outline of my remaining step 1. Show that $B,P,Q,C$ lie on the same circle. 2. Show that the circumcircle of $\triangle PQD$ touch $BC$ at $D$. 3. Comparing angles in the figure to show that $\angle KDA=\angle LDA$. Finally, we have $\triangle KDA \cong \triangle LDA$ and hence $KA=AL$. This implies that $KH \perp AD$. Sorry, perhaps I don't get how to prove 2.

jrpartty wrote: Extend $BH$ cut $AP$ at $Q$ and let $L$ be the intersection point between $AC$ and $QD$. We know that $K,H,L$ are collinear points from Pappus' theorem. Outline of my remaining step 1. Show that $B,P,Q,C$ lie on the same circle. 2. Show that the circumcircle of $\triangle PQD$ touch $BC$ at $D$. 3. Comparing angles in the figure to show that $\angle KDA=\angle LDA$. Finally, we have $\triangle KDA \cong \triangle LDA$ and hence $KA=AL$. This implies that $KH \perp AD$. Sorry but can you give the full solution please

egret_L wrote: jrpartty wrote: Extend $BH$ cut $AP$ at $Q$ and let $L$ be the intersection point between $AC$ and $QD$. We know that $K,H,L$ are collinear points from Pappus' theorem. Outline of my remaining step 1. Show that $B,P,Q,C$ lie on the same circle. 2. Show that the circumcircle of $\triangle PQD$ touch $BC$ at $D$. 3. Comparing angles in the figure to show that $\angle KDA=\angle LDA$. Finally, we have $\triangle KDA \cong \triangle LDA$ and hence $KA=AL$. This implies that $KH \perp AD$. Sorry, perhaps I don't get how to prove 2. Let $BC\cap PQ =S $ by angle chasing $SD = SA $ $ SD^2 = SA^2 = SB.SC = SP.SQ $