In $\triangle ABC$,$AB>AC$,let $H$ be $\triangle ABC$'s orthocenter,$M$ be $BC$'s midpoint.point $S$ is on $BC$ satisfies $\angle BHM=\angle CHS$.Point $P$ is on $HS$ so that $AP \perp HS$
Prove that the circumcircle of $\triangle MPS$ is tangent to the circumcircle of $\triangle ABC$
Circumcircle is tangent to the circumcircle
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Re: Circumcircle is tangent to the circumcircle
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