ATimo wrote: ABC is a triangle with A=90 and C=30.Let M be the midpoint of BC. Let W be a circle passing through A tangent in M to BC. Let P be the circumcircle of ABC. W is intersecting AC in N and P in M. prove that MN is perpendicular to BC. Let me correct your question and $\text{\LaTeX}\text{ify}$ it. Corrected QuestionQuestion $ABC$ is a triangle with $\angle BAC=90^{\circ}$ and $\angle ACB=30^{\circ}$. Let $M_1$ be the midpoint of $BC$. Let $W$ be a circle passing through $A$ tangent in $M_1$ to $BC$. Let $P$ be the circumcircle of $ABC$. $W$ is intersecting $AC$ in $N$ and $P$ in $M$. Prove that $MN$ is perpendicular to $BC$. SolutionSolution Let $M_1$ be the midpoint of $BC$. And let $C_1$ be the center of $W$. So, we first show that $C_1\in AC$. Clearly $C_1M_1\perp BC$. And also, since $M_1$ is the center of $P$ as $ABC$ is a right triangle, we have $M_1A=M_1B=M_1C$. So, $\angle M_1AB=\angle M_1BA=60^{\circ}$. And this implies that $\triangle AM_1B$ is equilateral. So, $\angle C_1M_1A=90^{\circ}-60^{\circ}=30^{\circ}=\angle C_1AM_1$. But then, $\angle C_1AM_1+\angle M_1AB=90^{\circ}=\angle C_1AB=\angle CAB$. This shows that $\boxed{C_1\in AC}$. Next $C_1N=C_1M_1$ and $\angle AC_1M_1=180^{\circ}-30^{\circ}-30^{\circ}=120^{\circ}=180^{\circ}-\angle NC_1M_1$, showing that $\triangle NC_1M_1$ is equilateral. Again, $C_1M=C_1A$ showing $\angle C_1AM=\angle C_1MA$. Now note that $AM$ is the radical axis of $P$ and $W$. So, $M_1C_1\perp AM$. Let $M_1C_1\cap AM=X$. Then, $AC_1X=60^{\circ}\Rightarrow NC_1M=180^{\circ}-\angle MC_1X-60^{\circ}=60^{\circ}$. But $C_1M=C_1N$. Thus $\angle MNC_1=\angle NC_1M\Rightarrow MN\parallel M_1C_1\Rightarrow \boxed{MN\perp BC}$. This completes the proof.

In the Quadrilateral $ABCD$ we have $ \measuredangle B=\measuredangle D = 60^\circ $.$M$ is midpoint of side $AD$.The line through $M$ parallel to $CD$ meets $BC$ at $P$.Point $X$ lying on $CD$ such that $BX=MX$.Prove that $AB=BP$ if and only if $\measuredangle MXB=60^\circ$. Author: Davoud Vakili, Iran

Let $ABC$ be an acute triangle.A circle with diameter $BC$ meets $AB$ and $AC$ at $E$ and $F$,respectively. $M$ is midpoint of $BC$ and $P$ is point of intersection $AM$ with $EF$. $X$ is an arbitary point on arc $EF$ and $Y$ is the second intersection of $XP$ with a circle with diameter $BC$.Prove that $ \measuredangle XAY=\measuredangle XYM $. Author:Ali zo'alam , Iran

A tangent line to circumcircle of acute triangle $ABC$ ($AC>AB$) at $A$ intersects with the extension of $BC$ at $P$. $O$ is the circumcenter of triangle $ABC$.Point $X$ lying on $OP$ such that $\measuredangle AXP=90^\circ$.Points $E$ and $F$ lying on $AB$ and $AC$,respectively,and they are in one side of line $OP$ such that $ \measuredangle EXP=\measuredangle ACX $ and $\measuredangle FXO=\measuredangle ABX $. $K$,$L$ are points of intersection $EF$ with circumcircle of triangle $ABC$.prove that $OP$ is tangent to circumcircle of triangle $KLX$. Author:Mehdi E'tesami Fard , Iran

Two points $P$ and $Q$ lying on side $BC$ of triangle $ABC$ and their distance from the midpoint of $BC$ are equal.The perpendiculars from $P$ and $Q$ to $BC$ intersect $AC$ and $AB$ at $E$ and $F$,respectively.$M$ is point of intersection $PF$ and $EQ$.If $H_1$ and $H_2$ be the orthocenters of triangles $BFP$ and $CEQ$, respectively, prove that $ AM\perp H_1H_2 $. Author:Mehdi E'tesami Fard , Iran