Vlados021 wrote: The point $M$ is the midpoint of the side $BC$ of triangle $ABC$. A circle is passing through $B$, is tangent to the line $AM$ at $M$, and intersects the segment $AB$ secondary at the point $P$. Prove that the circle, passing through $A$, $P$, and the midpoints of the segment $AM$, is tangent to the line $AC$. (A. Voidelevich) Let $D$ be a point such that $ABCD$ is a parallelogram and $E$ be midpoint of $AM$.Let $\odot{BMD}\cap AB=Q$.Since $AB\cdot AQ=AM\cdot AD=2AM^2 \Rightarrow AB\cdot \tfrac{AQ}{2}=AM^2$.Also since $AB\cdot AP=AM^2$,So $AQ=2AP$. Note that $\measuredangle MQA=\measuredangle MDA=\measuredangle MAC$.Hence $\odot{AMQ}$ is tangent to $AC$. Now a homothety $\mathbb{H}(A,2)$ maps $\odot{APE}\to \odot{AMQ}$. which finishes the proof.$\blacksquare$

Vlados021 wrote: The point $M$ is the midpoint of the side $BC$ of triangle $ABC$. A circle is passing through $B$, is tangent to the line $AM$ at $M$, and intersects the segment $AB$ secondary at the point $P$. Prove that the circle, passing through $A$, $P$, and the midpoints of the segment $AM$, is tangent to the line $AC$. Let $B'$ be reflection of $B$ in $A$, so $CB'||AM$ giving $\angle PAM = \angle BB'C$. Also, $\angle PMA = \angle CBB'$ giving $\triangle PAM\sim \triangle CB'B$. This gives that if $N$ is midpoint of $AM$, then $\triangle PNM\sim \triangle CAB$ which finishes the problem.

My solution: Let $N$ be the midpoint of side $AB$ and $O$ be the midpoint of median $AM$.Since the circumcircle of $BMP$ is tangent to the median we have $AM^2=AB\times AP$ .So dividing both sides by 2 we get that $AO\times AM=AN\times AB$ So $\quad NOBM$ is cyclic.(or we can use directly Reim's theorem) So $\angle POA=\angle MNA=180-A$.Therefore the circumcircle of $APO$ is tangent to segment $BC$

Let $Q$ be the midpoint of AM, $R$ be the midpoint of AB. Triangle $APM$ similar to $AMB$. Thus triangle $APQ$ similar to $AMR$. By midpoint theorem and the similarity, angle $AMR=C=APQ$ implying that $AC$ is tangent to the circumcircle of $APQ$.

Let $N$ be the midpoint of $AM$. Invert around $A$ with radius $AM$. Then $B$ and $P$ switch, and $N$ gets mapped to the reflection of $A$ over $M$. We wish to show that $AC^*$ and $BN^*$ are parallel, which is obvious since $ABN^* C$ is a parallelogram.

Let $N$, $Q$ be midpoint of $AB$, $AM$ Since: $AM$ tangents $(BPM)$ at $M$, we have: $\triangle ABM$ $\sim$ $\triangle AMP$ But: $N$, $Q$ are midpoint of $AB$, $AM$ then: $\triangle BNM$ $\sim$ $\triangle MQP$ So: $\angle{MQP} = \angle{BNM} = \angle{BAC}$ Hence: $\angle{AQP} = 180^o - \angle{BAC}$ or $AC$ tangents $(APQ)$ at $A$

Let $K$ - midpoint of $AM$. Let ($PMK$) meet $AB$ at $N$. By Reim's $KN \parallel BM$, so $N$ - midpoint of $AB$ and $MN \parallel AC$. Thus $\angle NPK = \angle NMK = \angle MAC$. $\blacksquare$