<KHC=<KCH=90-<ACB=90-<ABC and we are done.

easy! by alternate segment theorem we have $\angle ACH = \angle HPC$ where $P$ lie on major arc $HC$ thus $\angle HKC = 2\angle ACH = 2\angle ACB $ thus , $\angle HKC +\angle BAC = 180$ hence $\angle ADK = \angle ACK = 90 $ where $D$ is intersection of $HK,AB$ , since $ADKC$ is cyclic. and hence $KH$ is perpendicular to $AB$. and we are done !

Dear Mathlinkers, 1. (O) the circumcircle of ABC 2. Ta the tangent to (O) at A ; Ta // BC 3. N the second point of intersection of (O) with (K) 4. according to a spacial case of the Reim's theorem, A, H and N are collinear 5. Th the tangent to (K) at K 6. accordind to a spacial cas eof the Reim's theorem AB// Th and we are done... Sincerely Jean-Louis

Hi Here's my solution. We have that $ KHC=KCH=90-B=90-C $ So if $ KH $ and $ AB $ intersect at $D$ we have that $ DHB + B =90 $ so rhe problem is solved.

$AB$ is reflection of $AC$ in perpendicular bisector of BC, and so are the lines $HK$ and $KC$; since $KC\bot AC$, we are done. Remark: $H$ may be the second intersection of the circle $(K,KC)$ with the line $BC$. Best regards, sunken rock

Let <BAC= 2x and D intersection of KH and AB, then <ABC=<ACB= 90-x, and <BCK=x, hence <KHC=x=<BHD, and then <BDH=180-x-(90-x)=90

ComplexPhi wrote: Let $ABC$ be an isosceles triangle with $AB=AC$ . Let also $\omega$ be a circle of center $K$ tangent to the line $AC$ at $C$ which intersects the segment $BC$ again at $H$ . Prove that $HK \bot AB $. Let $KH$ intersect $AB$ in $F$.$\angle KCA = 90$.Let $\angle ABC = \angle ACB = x$ $\implies \angle BCK = 90 - x$ but $\angle CHK = \angle BHF = \angle HCK$ $\implies \angle BFH = 180 - x - (90 - x) = 90$.

Construct the tangent to $(K)$ at point $H$. Let it meet side $AC$ at point $M$. As the tangents from point $M$ on $(K)$ are $MH$ and $MC$, $MH=MC$. . $\implies \angle MHC=\angle MCH$ But, as $AB=AC$, $\angle ABC=\angle ACB=\angle MCH$.This implies that $MH \parallel AB$. Now as $MH$ is perpendicular to $KH$, $AB$ is perpendicular to $KH$. And we are done.

Let $D=KH\cap AB.$ Since $AB=AC$, $\angle B=\angle C$. Since $KC$ is tangent to $AC$, then $\angle ACK=90^{\circ}=\angle C+\angle BCK$. Since $KH=KC$, $\angle KCH=\angle KHC.$ Hence, $$\angle KDB=180^{\circ}-\angle B-\angle DHB= 180^{\circ}-\angle C-\angle KHC=180^{\circ}-\angle C-\angle KCB=90^{\circ},$$as desired. $\blacksquare$

just use the fact that K is the circumcenter and after that a simple angle chasing !

[asy][asy] import olympiad; size(10cm); pair A, B, C, K, H, P; A = (2, 6); B = (0, 0); C = (4, 0); draw(A--B--C--A); draw(circle((2, 2.67), sqrt(11.11))); K = (2.58, -0.47); draw(circle((2.58, -0.47), sqrt(2.24))); H = (1.16, 0); P = (0.12, 0.35); draw(K--P); draw(K--C); label("$A$", A, N); label("$B$", B, SW); label("$C$", C, SE); label("$K$", K, N); label("$P$", P, NW); label("$H$", H, SW); dot(A^^B^^C^^K^^H^^P); [/asy][/asy] Let $P$ be the intersection of line $HK$ and segment $AB$ and let $\angle{HKC}=\theta$. Then since $\triangle{KHC}$ is isosceles, we have $\angle{KHC}=\angle{KCH}=\angle{BHP}=90-\frac{\theta}{2}$. So then $\angle{ACB}=\angle{ABC}=\frac{\theta}{2}$. So then $\angle{BPH}=90^\circ$ and we are done.

Let KH meet AB at S. ∠CKS = 180 - 2∠HCK = 180 - 2(90 - ∠ACB) = 2∠ACB = 180 - ∠SAC ---> SACK is cyclic. ∠ACK + ∠ASK = 180 ---> ∠ASK = 90. we're Done.